Questions on binary search tree mirroring - binary-search-tree

I have written 2 methods to find out If an tree is an BST. Most these samples are from stanford courses on web.
before mirroring:
Tree is BST: true (results from first method)
Tree is BST2: true (results from second method)
after mirroring:
Tree is BST: false (results from first method)
Tree is BST2: true (results from second method)
If a binary search tree is mirrored. I believe it will not be an BST anymore. Is that right?
Is the second method wrong?
//mirror
public void mirror(){
mirror(root);
}
private void mirror(Node node){
if(node == null) return;
if( node != null){
mirror(node.left);
mirror(node.right);
Node temp = node.left;
node.left = node.right;
node.right = temp;
}
}
public boolean isBST(){
return isBST(root);
}
private boolean isBST(Node node){
if( node == null) return true;
if( node.left != null && maxValue(node.left) > node.data) return false;
if( node.right != null && minValue(node.right) <= node.data) return false;
return (isBST(node.left) && isBST(node.right));
}
public boolean isBST2(){
return isBST2(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
private boolean isBST2(Node node, int minVal, int maxVal){
if( node == null) return true;
else{
boolean leftOk = isBST2(node.left, minVal, node.data);
if(!leftOk) return false;
boolean rightOk = isBST2(node.right, node.data+1, maxVal);
return rightOk;
}
}
//insert with inserts - tree increases on right if inserted in order
bt = new BinaryTree();
bt.insert(5);
bt.insert(3);
bt.insert(7);
bt.insert(1);
bt.insert(4);
bt.insert(6);
bt.insert(9);
bt.printTree();
bt.printPostorder();
System.out.println("max depth: " + bt.maxDepth());
System.out.println("min value: " + bt.minValue());
System.out.println("max value: " + bt.maxValue());
System.out.println("size of tree: " + bt.size());
System.out.println("Has path sum 8: " + bt.hasPathSum(8));
System.out.println("Has path sum 9: " + bt.hasPathSum(9));
bt.printPath();
System.out.println("Trees are same: " + bt.sameTree(bt));
System.out.println("Trees is BST: " + bt.isBST());
System.out.println("Trees is BST: " + bt.isBST2());
bt.mirror();
System.out.println("Path after mirroring BST");
bt.printPath();
System.out.println("Trees is BST: " + bt.isBST());
System.out.println("Trees is BST: " + bt.isBST2());
System.out.println("---------------------");
}

isBST2() does not correctly return if the tree is a BST or not.
In fact it always returns true. When node is null it will return true. Further see that in isBST2() there is no condition checking except if(!leftOk). So if leftOk is true it does not return false. So the function always returns true. Hence this function is not correct.
Whereas, the function isBST() is correct.
So your guess is correct.
Also, when a BST is mirrored, it does not remain a BST, unless it contains only one node.

Related

Arithmetic parser in kotlin [duplicate]

I'm trying to write a Java routine to evaluate math expressions from String values like:
"5+3"
"10-4*5"
"(1+10)*3"
I want to avoid a lot of if-then-else statements.
How can I do this?
With JDK1.6, you can use the built-in Javascript engine.
import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;
import javax.script.ScriptException;
public class Test {
public static void main(String[] args) throws ScriptException {
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine engine = mgr.getEngineByName("JavaScript");
String foo = "40+2";
System.out.println(engine.eval(foo));
}
}
I've written this eval method for arithmetic expressions to answer this question. It does addition, subtraction, multiplication, division, exponentiation (using the ^ symbol), and a few basic functions like sqrt. It supports grouping using (...), and it gets the operator precedence and associativity rules correct.
public static double eval(final String str) {
return new Object() {
int pos = -1, ch;
void nextChar() {
ch = (++pos < str.length()) ? str.charAt(pos) : -1;
}
boolean eat(int charToEat) {
while (ch == ' ') nextChar();
if (ch == charToEat) {
nextChar();
return true;
}
return false;
}
double parse() {
nextChar();
double x = parseExpression();
if (pos < str.length()) throw new RuntimeException("Unexpected: " + (char)ch);
return x;
}
// Grammar:
// expression = term | expression `+` term | expression `-` term
// term = factor | term `*` factor | term `/` factor
// factor = `+` factor | `-` factor | `(` expression `)` | number
// | functionName `(` expression `)` | functionName factor
// | factor `^` factor
double parseExpression() {
double x = parseTerm();
for (;;) {
if (eat('+')) x += parseTerm(); // addition
else if (eat('-')) x -= parseTerm(); // subtraction
else return x;
}
}
double parseTerm() {
double x = parseFactor();
for (;;) {
if (eat('*')) x *= parseFactor(); // multiplication
else if (eat('/')) x /= parseFactor(); // division
else return x;
}
}
double parseFactor() {
if (eat('+')) return +parseFactor(); // unary plus
if (eat('-')) return -parseFactor(); // unary minus
double x;
int startPos = this.pos;
if (eat('(')) { // parentheses
x = parseExpression();
if (!eat(')')) throw new RuntimeException("Missing ')'");
} else if ((ch >= '0' && ch <= '9') || ch == '.') { // numbers
while ((ch >= '0' && ch <= '9') || ch == '.') nextChar();
x = Double.parseDouble(str.substring(startPos, this.pos));
} else if (ch >= 'a' && ch <= 'z') { // functions
while (ch >= 'a' && ch <= 'z') nextChar();
String func = str.substring(startPos, this.pos);
if (eat('(')) {
x = parseExpression();
if (!eat(')')) throw new RuntimeException("Missing ')' after argument to " + func);
} else {
x = parseFactor();
}
if (func.equals("sqrt")) x = Math.sqrt(x);
else if (func.equals("sin")) x = Math.sin(Math.toRadians(x));
else if (func.equals("cos")) x = Math.cos(Math.toRadians(x));
else if (func.equals("tan")) x = Math.tan(Math.toRadians(x));
else throw new RuntimeException("Unknown function: " + func);
} else {
throw new RuntimeException("Unexpected: " + (char)ch);
}
if (eat('^')) x = Math.pow(x, parseFactor()); // exponentiation
return x;
}
}.parse();
}
Example:
System.out.println(eval("((4 - 2^3 + 1) * -sqrt(3*3+4*4)) / 2"));
Output: 7.5 (which is correct)
The parser is a recursive descent parser, so internally uses separate parse methods for each level of operator precedence in its grammar. I deliberately kept it short, but here are some ideas you might want to expand it with:
Variables:
The bit of the parser that reads the names for functions can easily be changed to handle custom variables too, by looking up names in a variable table passed to the eval method, such as a Map<String,Double> variables.
Separate compilation and evaluation:
What if, having added support for variables, you wanted to evaluate the same expression millions of times with changed variables, without parsing it every time? It's possible. First define an interface to use to evaluate the precompiled expression:
#FunctionalInterface
interface Expression {
double eval();
}
Now to rework the original "eval" function into a "parse" function, change all the methods that return doubles, so instead they return an instance of that interface. Java 8's lambda syntax works well for this. Example of one of the changed methods:
Expression parseExpression() {
Expression x = parseTerm();
for (;;) {
if (eat('+')) { // addition
Expression a = x, b = parseTerm();
x = (() -> a.eval() + b.eval());
} else if (eat('-')) { // subtraction
Expression a = x, b = parseTerm();
x = (() -> a.eval() - b.eval());
} else {
return x;
}
}
}
That builds a recursive tree of Expression objects representing the compiled expression (an abstract syntax tree). Then you can compile it once and evaluate it repeatedly with different values:
public static void main(String[] args) {
Map<String,Double> variables = new HashMap<>();
Expression exp = parse("x^2 - x + 2", variables);
for (double x = -20; x <= +20; x++) {
variables.put("x", x);
System.out.println(x + " => " + exp.eval());
}
}
Different datatypes:
Instead of double, you could change the evaluator to use something more powerful like BigDecimal, or a class that implements complex numbers, or rational numbers (fractions). You could even use Object, allowing some mix of datatypes in expressions, just like a real programming language. :)
All code in this answer released to the public domain. Have fun!
For my university project, I was looking for a parser / evaluator supporting both basic formulas and more complicated equations (especially iterated operators). I found very nice open source library for JAVA and .NET called mXparser. I will give a few examples to make some feeling on the syntax, for further instructions please visit project website (especially tutorial section).
https://mathparser.org/
https://mathparser.org/mxparser-tutorial/
https://mathparser.org/api/
And few examples
1 - Simple furmula
Expression e = new Expression("( 2 + 3/4 + sin(pi) )/2");
double v = e.calculate()
2 - User defined arguments and constants
Argument x = new Argument("x = 10");
Constant a = new Constant("a = pi^2");
Expression e = new Expression("cos(a*x)", x, a);
double v = e.calculate()
3 - User defined functions
Function f = new Function("f(x, y, z) = sin(x) + cos(y*z)");
Expression e = new Expression("f(3,2,5)", f);
double v = e.calculate()
4 - Iteration
Expression e = new Expression("sum( i, 1, 100, sin(i) )");
double v = e.calculate()
Found recently - in case you would like to try the syntax (and see the advanced use case) you can download the Scalar Calculator app that is powered by mXparser.
The correct way to solve this is with a lexer and a parser. You can write simple versions of these yourself, or those pages also have links to Java lexers and parsers.
Creating a recursive descent parser is a really good learning exercise.
HERE is another open source library on GitHub named EvalEx.
Unlike the JavaScript engine this library is focused in evaluating mathematical expressions only. Moreover, the library is extensible and supports use of boolean operators as well as parentheses.
You can evaluate expressions easily if your Java application already accesses a database, without using any other JARs.
Some databases require you to use a dummy table (eg, Oracle's "dual" table) and others will allow you to evaluate expressions without "selecting" from any table.
For example, in Sql Server or Sqlite
select (((12.10 +12.0))/ 233.0) amount
and in Oracle
select (((12.10 +12.0))/ 233.0) amount from dual;
The advantage of using a DB is that you can evaluate many expressions at the same time. Also most DB's will allow you to use highly complex expressions and will also have a number of extra functions that can be called as necessary.
However performance may suffer if many single expressions need to be evaluated individually, particularly when the DB is located on a network server.
The following addresses the performance problem to some extent, by using a Sqlite in-memory database.
Here's a full working example in Java
Class. forName("org.sqlite.JDBC");
Connection conn = DriverManager.getConnection("jdbc:sqlite::memory:");
Statement stat = conn.createStatement();
ResultSet rs = stat.executeQuery( "select (1+10)/20.0 amount");
rs.next();
System.out.println(rs.getBigDecimal(1));
stat.close();
conn.close();
Of course you could extend the above code to handle multiple calculations at the same time.
ResultSet rs = stat.executeQuery( "select (1+10)/20.0 amount, (1+100)/20.0 amount2");
You can also try the BeanShell interpreter:
Interpreter interpreter = new Interpreter();
interpreter.eval("result = (7+21*6)/(32-27)");
System.out.println(interpreter.get("result"));
Another way is to use the Spring Expression Language or SpEL which does a whole lot more along with evaluating mathematical expressions, therefore maybe slightly overkill. You do not have to be using Spring framework to use this expression library as it is stand-alone. Copying examples from SpEL's documentation:
ExpressionParser parser = new SpelExpressionParser();
int two = parser.parseExpression("1 + 1").getValue(Integer.class); // 2
double twentyFour = parser.parseExpression("2.0 * 3e0 * 4").getValue(Double.class); //24.0
This article discusses various approaches. Here are the 2 key approaches mentioned in the article:
JEXL from Apache
Allows for scripts that include references to java objects.
// Create or retrieve a JexlEngine
JexlEngine jexl = new JexlEngine();
// Create an expression object
String jexlExp = "foo.innerFoo.bar()";
Expression e = jexl.createExpression( jexlExp );
// Create a context and add data
JexlContext jctx = new MapContext();
jctx.set("foo", new Foo() );
// Now evaluate the expression, getting the result
Object o = e.evaluate(jctx);
Use the javascript engine embedded in the JDK:
private static void jsEvalWithVariable()
{
List<String> namesList = new ArrayList<String>();
namesList.add("Jill");
namesList.add("Bob");
namesList.add("Laureen");
namesList.add("Ed");
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine jsEngine = mgr.getEngineByName("JavaScript");
jsEngine.put("namesListKey", namesList);
System.out.println("Executing in script environment...");
try
{
jsEngine.eval("var x;" +
"var names = namesListKey.toArray();" +
"for(x in names) {" +
" println(names[x]);" +
"}" +
"namesListKey.add(\"Dana\");");
}
catch (ScriptException ex)
{
ex.printStackTrace();
}
}
if we are going to implement it then we can can use the below algorithm :--
While there are still tokens to be read in,
1.1 Get the next token.
1.2 If the token is:
1.2.1 A number: push it onto the value stack.
1.2.2 A variable: get its value, and push onto the value stack.
1.2.3 A left parenthesis: push it onto the operator stack.
1.2.4 A right parenthesis:
1 While the thing on top of the operator stack is not a
left parenthesis,
1 Pop the operator from the operator stack.
2 Pop the value stack twice, getting two operands.
3 Apply the operator to the operands, in the correct order.
4 Push the result onto the value stack.
2 Pop the left parenthesis from the operator stack, and discard it.
1.2.5 An operator (call it thisOp):
1 While the operator stack is not empty, and the top thing on the
operator stack has the same or greater precedence as thisOp,
1 Pop the operator from the operator stack.
2 Pop the value stack twice, getting two operands.
3 Apply the operator to the operands, in the correct order.
4 Push the result onto the value stack.
2 Push thisOp onto the operator stack.
While the operator stack is not empty,
1 Pop the operator from the operator stack.
2 Pop the value stack twice, getting two operands.
3 Apply the operator to the operands, in the correct order.
4 Push the result onto the value stack.
At this point the operator stack should be empty, and the value
stack should have only one value in it, which is the final result.
This is another interesting alternative
https://github.com/Shy-Ta/expression-evaluator-demo
The usage is very simple and gets the job done, for example:
ExpressionsEvaluator evalExpr = ExpressionsFactory.create("2+3*4-6/2");
assertEquals(BigDecimal.valueOf(11), evalExpr.eval());
It seems like JEP should do the job
It's too late to answer but I came across same situation to evaluate expression in java, it might help someone
MVEL does runtime evaluation of expressions, we can write a java code in String to get it evaluated in this.
String expressionStr = "x+y";
Map<String, Object> vars = new HashMap<String, Object>();
vars.put("x", 10);
vars.put("y", 20);
ExecutableStatement statement = (ExecutableStatement) MVEL.compileExpression(expressionStr);
Object result = MVEL.executeExpression(statement, vars);
Try the following sample code using JDK1.6's Javascript engine with code injection handling.
import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
public class EvalUtil {
private static ScriptEngine engine = new ScriptEngineManager().getEngineByName("JavaScript");
public static void main(String[] args) {
try {
System.out.println((new EvalUtil()).eval("(((5+5)/2) > 5) || 5 >3 "));
System.out.println((new EvalUtil()).eval("(((5+5)/2) > 5) || true"));
} catch (Exception e) {
e.printStackTrace();
}
}
public Object eval(String input) throws Exception{
try {
if(input.matches(".*[a-zA-Z;~`#$_{}\\[\\]:\\\\;\"',\\.\\?]+.*")) {
throw new Exception("Invalid expression : " + input );
}
return engine.eval(input);
} catch (Exception e) {
e.printStackTrace();
throw e;
}
}
}
This is actually complementing the answer given by #Boann. It has a slight bug which causes "-2 ^ 2" to give an erroneous result of -4.0. The problem for that is the point at which the exponentiation is evaluated in his. Just move the exponentiation to the block of parseTerm(), and you'll be all fine. Have a look at the below, which is #Boann's answer slightly modified. Modification is in the comments.
public static double eval(final String str) {
return new Object() {
int pos = -1, ch;
void nextChar() {
ch = (++pos < str.length()) ? str.charAt(pos) : -1;
}
boolean eat(int charToEat) {
while (ch == ' ') nextChar();
if (ch == charToEat) {
nextChar();
return true;
}
return false;
}
double parse() {
nextChar();
double x = parseExpression();
if (pos < str.length()) throw new RuntimeException("Unexpected: " + (char)ch);
return x;
}
// Grammar:
// expression = term | expression `+` term | expression `-` term
// term = factor | term `*` factor | term `/` factor
// factor = `+` factor | `-` factor | `(` expression `)`
// | number | functionName factor | factor `^` factor
double parseExpression() {
double x = parseTerm();
for (;;) {
if (eat('+')) x += parseTerm(); // addition
else if (eat('-')) x -= parseTerm(); // subtraction
else return x;
}
}
double parseTerm() {
double x = parseFactor();
for (;;) {
if (eat('*')) x *= parseFactor(); // multiplication
else if (eat('/')) x /= parseFactor(); // division
else if (eat('^')) x = Math.pow(x, parseFactor()); //exponentiation -> Moved in to here. So the problem is fixed
else return x;
}
}
double parseFactor() {
if (eat('+')) return parseFactor(); // unary plus
if (eat('-')) return -parseFactor(); // unary minus
double x;
int startPos = this.pos;
if (eat('(')) { // parentheses
x = parseExpression();
eat(')');
} else if ((ch >= '0' && ch <= '9') || ch == '.') { // numbers
while ((ch >= '0' && ch <= '9') || ch == '.') nextChar();
x = Double.parseDouble(str.substring(startPos, this.pos));
} else if (ch >= 'a' && ch <= 'z') { // functions
while (ch >= 'a' && ch <= 'z') nextChar();
String func = str.substring(startPos, this.pos);
x = parseFactor();
if (func.equals("sqrt")) x = Math.sqrt(x);
else if (func.equals("sin")) x = Math.sin(Math.toRadians(x));
else if (func.equals("cos")) x = Math.cos(Math.toRadians(x));
else if (func.equals("tan")) x = Math.tan(Math.toRadians(x));
else throw new RuntimeException("Unknown function: " + func);
} else {
throw new RuntimeException("Unexpected: " + (char)ch);
}
//if (eat('^')) x = Math.pow(x, parseFactor()); // exponentiation -> This is causing a bit of problem
return x;
}
}.parse();
}
import java.util.*;
public class check {
int ans;
String str="7 + 5";
StringTokenizer st=new StringTokenizer(str);
int v1=Integer.parseInt(st.nextToken());
String op=st.nextToken();
int v2=Integer.parseInt(st.nextToken());
if(op.equals("+")) { ans= v1 + v2; }
if(op.equals("-")) { ans= v1 - v2; }
//.........
}
I think what ever way you do this it's going to involve a lot of conditional statements. But for single operations like in your examples you could limit it to 4 if statements with something like
String math = "1+4";
if (math.split("+").length == 2) {
//do calculation
} else if (math.split("-").length == 2) {
//do calculation
} ...
It gets a whole lot more complicated when you want to deal with multiple operations like "4+5*6".
If you are trying to build a calculator then I'd surgest passing each section of the calculation separatly (each number or operator) rather than as a single string.
You might have a look at the Symja framework:
ExprEvaluator util = new ExprEvaluator();
IExpr result = util.evaluate("10-40");
System.out.println(result.toString()); // -> "-30"
Take note that definitively more complex expressions can be evaluated:
// D(...) gives the derivative of the function Sin(x)*Cos(x)
IAST function = D(Times(Sin(x), Cos(x)), x);
IExpr result = util.evaluate(function);
// print: Cos(x)^2-Sin(x)^2
package ExpressionCalculator.expressioncalculator;
import java.text.DecimalFormat;
import java.util.Scanner;
public class ExpressionCalculator {
private static String addSpaces(String exp){
//Add space padding to operands.
//https://regex101.com/r/sJ9gM7/73
exp = exp.replaceAll("(?<=[0-9()])[\\/]", " / ");
exp = exp.replaceAll("(?<=[0-9()])[\\^]", " ^ ");
exp = exp.replaceAll("(?<=[0-9()])[\\*]", " * ");
exp = exp.replaceAll("(?<=[0-9()])[+]", " + ");
exp = exp.replaceAll("(?<=[0-9()])[-]", " - ");
//Keep replacing double spaces with single spaces until your string is properly formatted
/*while(exp.indexOf(" ") != -1){
exp = exp.replace(" ", " ");
}*/
exp = exp.replaceAll(" {2,}", " ");
return exp;
}
public static Double evaluate(String expr){
DecimalFormat df = new DecimalFormat("#.####");
//Format the expression properly before performing operations
String expression = addSpaces(expr);
try {
//We will evaluate using rule BDMAS, i.e. brackets, division, power, multiplication, addition and
//subtraction will be processed in following order
int indexClose = expression.indexOf(")");
int indexOpen = -1;
if (indexClose != -1) {
String substring = expression.substring(0, indexClose);
indexOpen = substring.lastIndexOf("(");
substring = substring.substring(indexOpen + 1).trim();
if(indexOpen != -1 && indexClose != -1) {
Double result = evaluate(substring);
expression = expression.substring(0, indexOpen).trim() + " " + result + " " + expression.substring(indexClose + 1).trim();
return evaluate(expression.trim());
}
}
String operation = "";
if(expression.indexOf(" / ") != -1){
operation = "/";
}else if(expression.indexOf(" ^ ") != -1){
operation = "^";
} else if(expression.indexOf(" * ") != -1){
operation = "*";
} else if(expression.indexOf(" + ") != -1){
operation = "+";
} else if(expression.indexOf(" - ") != -1){ //Avoid negative numbers
operation = "-";
} else{
return Double.parseDouble(expression);
}
int index = expression.indexOf(operation);
if(index != -1){
indexOpen = expression.lastIndexOf(" ", index - 2);
indexOpen = (indexOpen == -1)?0:indexOpen;
indexClose = expression.indexOf(" ", index + 2);
indexClose = (indexClose == -1)?expression.length():indexClose;
if(indexOpen != -1 && indexClose != -1) {
Double lhs = Double.parseDouble(expression.substring(indexOpen, index));
Double rhs = Double.parseDouble(expression.substring(index + 2, indexClose));
Double result = null;
switch (operation){
case "/":
//Prevent divide by 0 exception.
if(rhs == 0){
return null;
}
result = lhs / rhs;
break;
case "^":
result = Math.pow(lhs, rhs);
break;
case "*":
result = lhs * rhs;
break;
case "-":
result = lhs - rhs;
break;
case "+":
result = lhs + rhs;
break;
default:
break;
}
if(indexClose == expression.length()){
expression = expression.substring(0, indexOpen) + " " + result + " " + expression.substring(indexClose);
}else{
expression = expression.substring(0, indexOpen) + " " + result + " " + expression.substring(indexClose + 1);
}
return Double.valueOf(df.format(evaluate(expression.trim())));
}
}
}catch(Exception exp){
exp.printStackTrace();
}
return 0.0;
}
public static void main(String args[]){
Scanner scanner = new Scanner(System.in);
System.out.print("Enter an Mathematical Expression to Evaluate: ");
String input = scanner.nextLine();
System.out.println(evaluate(input));
}
}
A Java class that can evaluate mathematical expressions:
package test;
public class Calculator {
public static Double calculate(String expression){
if (expression == null || expression.length() == 0) {
return null;
}
return calc(expression.replace(" ", ""));
}
public static Double calc(String expression) {
String[] containerArr = new String[]{expression};
double leftVal = getNextOperand(containerArr);
expression = containerArr[0];
if (expression.length() == 0) {
return leftVal;
}
char operator = expression.charAt(0);
expression = expression.substring(1);
while (operator == '*' || operator == '/') {
containerArr[0] = expression;
double rightVal = getNextOperand(containerArr);
expression = containerArr[0];
if (operator == '*') {
leftVal = leftVal * rightVal;
} else {
leftVal = leftVal / rightVal;
}
if (expression.length() > 0) {
operator = expression.charAt(0);
expression = expression.substring(1);
} else {
return leftVal;
}
}
if (operator == '+') {
return leftVal + calc(expression);
} else {
return leftVal - calc(expression);
}
}
private static double getNextOperand(String[] exp){
double res;
if (exp[0].startsWith("(")) {
int open = 1;
int i = 1;
while (open != 0) {
if (exp[0].charAt(i) == '(') {
open++;
} else if (exp[0].charAt(i) == ')') {
open--;
}
i++;
}
res = calc(exp[0].substring(1, i - 1));
exp[0] = exp[0].substring(i);
} else {
int i = 1;
if (exp[0].charAt(0) == '-') {
i++;
}
while (exp[0].length() > i && isNumber((int) exp[0].charAt(i))) {
i++;
}
res = Double.parseDouble(exp[0].substring(0, i));
exp[0] = exp[0].substring(i);
}
return res;
}
private static boolean isNumber(int c) {
int zero = (int) '0';
int nine = (int) '9';
return (c >= zero && c <= nine) || c =='.';
}
public static void main(String[] args) {
System.out.println(calculate("(((( -6 )))) * 9 * -1"));
System.out.println(calc("(-5.2+-5*-5*((5/4+2)))"));
}
}
How about something like this:
String st = "10+3";
int result;
for(int i=0;i<st.length();i++)
{
if(st.charAt(i)=='+')
{
result=Integer.parseInt(st.substring(0, i))+Integer.parseInt(st.substring(i+1, st.length()));
System.out.print(result);
}
}
and do the similar thing for every other mathematical operator accordingly ..
It is possible to convert any expression string in infix notation to a postfix notation using Djikstra's shunting-yard algorithm. The result of the algorithm can then serve as input to the postfix algorithm with returns the result of the expression.
I wrote an article about it here, with an implementation in java
Yet another option: https://github.com/stefanhaustein/expressionparser
I have implemented this to have a simple but flexible option to permit both:
Immediate processing (Calculator.java, SetDemo.java)
Building and processing a parse tree (TreeBuilder.java)
The TreeBuilder linked above is part of a CAS demo package that does symbolic derivation. There is also a BASIC interpreter example and I have started to build a TypeScript interpreter using it.
External library like RHINO or NASHORN can be used to run javascript. And javascript can evaluate simple formula without parcing the string. No performance impact as well if code is written well.
Below is an example with RHINO -
public class RhinoApp {
private String simpleAdd = "(12+13+2-2)*2+(12+13+2-2)*2";
public void runJavaScript() {
Context jsCx = Context.enter();
Context.getCurrentContext().setOptimizationLevel(-1);
ScriptableObject scope = jsCx.initStandardObjects();
Object result = jsCx.evaluateString(scope, simpleAdd , "formula", 0, null);
Context.exit();
System.out.println(result);
}
import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
import javax.script.ScriptException;
public class test2 {
public static void main(String[] args) throws ScriptException {
String s = "10+2";
ScriptEngineManager mn = new ScriptEngineManager();
ScriptEngine en = mn.getEngineByName("js");
Object result = en.eval(s);
System.out.println(result);
}
}
I have done using iterative parsing and shunting Yard algorithm and i have really enjoyed developing the expression evaluator ,you can find all the code here
https://github.com/nagaraj200788/JavaExpressionEvaluator
Has 73 test cases and even works for Bigintegers,Bigdecimals
supports all relational, arithmetic expression and also combination of both .
even supports ternary operator .
Added enhancement to support signed numbers like -100+89 it was intresting, for details check TokenReader.isUnaryOperator() method and i have updated code in above Link

Is there any method in ByteBuddy to convert a TypeDescription.Generic into an appropriate java.lang.reflect.Type?

(The surface area of the ByteBuddy API is overwhelmingly enormous, which is why I'm asking the question.)
I'm aware that I can take a TypeDescription.Generic and determine its "sort" and proceed rather laboriously "by hand" from there, but often times I've found there is a method buried somewhere that will do this sort of tedious work for me.
EDIT: a commenter asked for the "tedious" recipe. Here it is (stand back; please note the various implementations of various Types are more or less what you'd expect them to be):
public static final Type toType(final TypeDefinition type) throws ReflectiveOperationException {
final Type returnValue;
if (type == null) {
returnValue = null;
} else {
final TypeDescription.Generic genericType = type.asGenericType();
switch (type.getSort()) {
case GENERIC_ARRAY:
returnValue = new DefaultGenericArrayType(toType(type.getComponentType()));
break;
case NON_GENERIC:
returnValue = Class.forName(type.getTypeName(), false, Thread.currentThread().getContextClassLoader());
break;
case PARAMETERIZED:
final TypeDefinition ownerType = genericType.getOwnerType();
final TypeDefinition rawType = type.asErasure();
final List<? extends TypeDefinition> actualTypeArguments = genericType.getTypeArguments();
if (actualTypeArguments == null || actualTypeArguments.isEmpty()) {
returnValue = new DefaultParameterizedType(toType(ownerType), toType(rawType));
} else {
final Type[] actualJavaTypeArguments = new Type[actualTypeArguments.size()];
for (int i = 0; i < actualTypeArguments.size(); i++) {
actualJavaTypeArguments[i] = toType(actualTypeArguments.get(i));
}
returnValue = new DefaultParameterizedType(toType(ownerType), toType(rawType), actualJavaTypeArguments);
}
break;
case VARIABLE:
final TypeVariableSource typeVariableSource = genericType.getTypeVariableSource();
final GenericDeclaration gd;
if (typeVariableSource instanceof TypeDefinition typeDefinition) {
gd = Class.forName(typeDefinition.asErasure().getTypeName(), false, Thread.currentThread().getContextClassLoader());
} else if (typeVariableSource instanceof MethodDescription.InDefinedShape methodDescription) {
// Reflection time
final String name = methodDescription.getName();
final Class<?> cls = Class.forName(methodDescription.getDeclaringType().asErasure().getTypeName(), false, Thread.currentThread().getContextClassLoader());
final List<? extends TypeDefinition> parameterTypes = methodDescription.getParameters().asTypeList();
final Class<?>[] parameterClasses = new Class<?>[parameterTypes.size()];
for (int i = 0; i < parameterTypes.size(); i++) {
parameterClasses[i] = Class.forName(parameterTypes.get(i).asErasure().getName(), false, Thread.currentThread().getContextClassLoader());
}
if (MethodDescription.CONSTRUCTOR_INTERNAL_NAME.equals(name)) {
assert TypeDescription.VOID.equals(methodDescription.getReturnType());
gd = cls.getDeclaredConstructor(parameterClasses);
} else {
assert !MethodDescription.TYPE_INITIALIZER_INTERNAL_NAME.equals(name);
gd = cls.getDeclaredMethod(name, parameterClasses);
}
} else {
throw new IllegalArgumentException("Unexpected type variable source: " + typeVariableSource);
}
final TypeVariable<?>[] typeVariables = gd.getTypeParameters();
TypeVariable<?> temp = null;
for (final TypeVariable<?> typeVariable : typeVariables) {
if (typeVariable.getName().equals(genericType.getSymbol())) {
temp = typeVariable;
break;
}
}
assert temp != null;
returnValue = temp;
break;
case VARIABLE_SYMBOLIC:
throw new IllegalArgumentException("Unexpected type: " + type);
case WILDCARD:
final List<? extends TypeDefinition> upperBounds = genericType.getUpperBounds();
final List<? extends TypeDefinition> lowerBounds = genericType.getLowerBounds();
if (lowerBounds == null || lowerBounds.isEmpty()) {
if (upperBounds == null || upperBounds.isEmpty() || (upperBounds.size() == 1 && TypeDescription.Generic.OBJECT.equals(upperBounds.get(0)))) {
returnValue = UnboundedWildcardType.INSTANCE;
} else {
// Upper bounded.
final Type[] upperJavaBounds = new Type[upperBounds.size()];
for (int i = 0; i < upperBounds.size(); i++) {
upperJavaBounds[i] = toType(upperBounds.get(i)); // XXX recursive
}
returnValue = new UpperBoundedWildcardType(upperJavaBounds);
}
} else {
assert upperBounds == null || upperBounds.isEmpty() || (upperBounds.size() == 1 && TypeDescription.Generic.OBJECT.equals(upperBounds.get(0))) : "Unexpected upper bounds: " + upperBounds + "; lower bounds: " + lowerBounds;
// Lower bounded.
assert lowerBounds.size() == 1 : "Unexpected size in lower bounds: " + lowerBounds;
returnValue = new LowerBoundedWildcardType(toType(lowerBounds.get(0))); // XXX recursive
}
break;
default:
throw new IllegalArgumentException("Unexpected type: " + type);
}
}
return returnValue;
}
No, you can only convert a Type to a TypeDescription.Generic but there is no option to do it the other way. The easiest option to emulate this would probably be to define a class that defines a field of the given Type, to load this class and to read the field type using Java reflection.
The reason Byte Buddy cannot convert a description to a Type is that Byte Buddy abstracts out class loaders and that type variables might be detached from their declaring source.

BST, is it possible to find next lowest in O(lg N)?

The TreeNode class is defined with only left and right child.
public class TreeNode {
public int val;
public TreeNode left, right;
public TreeNode(int val) {
this.val = val;
}
}
My code finds the next lowest node in O(n). I was wondering if it's possible to find it in lg(N) given that the node doesn't have a pointer to its parent node.
// run time O(n)
public static Integer findNextLowest(TreeNode root, int target) {
Stack<TreeNode> stack = new Stack<>();
TreeNode cur = root;
while (cur != null || stack.size() > 0) {
while (cur != null) {
stack.push(cur);
cur = cur.right;
}
TreeNode node = stack.pop();
if (node.val < target) return node.val; // found the next lowest
cur = node.left;
}
return null;
}
private static TreeNode findNextLowest(TreeNode root, int target){
TreeNode node = root;
TreeNode res = null;
while(node != null){
while(node != null && node.val >= target){
node = node.left;
}
while(node != null && node.val < target){
res = node;
node = node.right;
}
}
return res;
}
No, because you haven't implemented a Binary Search Tree, just a Binary Tree.
A BST will constrain its values such that left.val < val < right.val, so you can do
// run time O(log(n)) if cur is balanced
public static Integer findNextLowest(TreeNode cur, int target) {
if (target < cur.val) { return cur.left != null ? findNextLowest(cur.left, target) : null; }
if (curr.right != null)
{
Integer result = findNextLowest(cur.right, target);
if (result != null) { return result; }
}
return cur.val;
}
You should use something like a R-B tree to ensure it is balanced

Deleting an node in BST

This is not an homework. I am just totally blocked on this. I know what to do but I am having difficulty manipulating the tree. please help.
I am trying to delete and node from an BST. I am able to lookup and find the parent and store it an tree.
package com.test.binarytree;
public class BinaryTreeDelete {
private Node root;
//create null binary tree
public BinaryTreeDelete(){
root = null;
}
//delete
public void delete(int target){
root = delete(root, target);
}
public Node delete(Node node, int target){
NodeWithParent temp = lookupFindParent(root, null, target);
if( node == null){
return null;
}
else{
if( node.left == null || node.right == null) //leaf node
{
//WHAT DO I DO HERE
//temp.parent.left = null;
//temp.parent.right = null;
//return null;
}
if( node.left != null && node.right == null ) //one child only on left
{
//WHAT DO I DO HERE
}
if( node.right != null && node.left == null ) //one child only on right
{
//WHAT DO I DO HERE
}
if( node.left != null && node.right != null ) //two children
{
//WHAT DO I DO HERE
}
}
return null;
}
private NodeWithParent lookupFindParent(Node node, Node parentNode, int target){
if( node == null ){
return null;
}
if( node.data == target){
return new NodeWithParent(node, parentNode);
}
else if( node.data > target ){
parentNode = node;
return lookupFindParent(node.left, parentNode, target);
}
else{
parentNode = node;
return lookupFindParent(node.right, parentNode, target);
}
}
//insert
public void insert(int data){
root = insert(root, data);
}
public Node insert (Node node, int data){
if(node == null){
node = new Node(data);
}
else{
if( data <= node.data ){
node.left = insert(node.left, data);
}
else{
node.right = insert(node.right, data);
}
}
return node;
}
//print tree
public void printTree(){
printTree(root);
System.out.println();
}
//print tree
private void printTree(Node node) {
if (node == null) return;
// left, node itself, right
printTree(node.left);
System.out.print(node.data + " ");
printTree(node.right);
}
//node class
public static class Node{
Node left;
Node right;
int data;
Node(int newNode){
data = newNode;
left = null;
right = null;
}
}
//node class
public static class NodeWithParent{
Node current;
Node parent;
NodeWithParent(Node current, Node parent){
this.current = current;
this.parent = parent;
}
}
public static void main(String[] args) {
BinaryTreeDelete bt = new BinaryTreeDelete();
//insert with inserts - tree increases on right if inserted in order
bt = new BinaryTreeDelete();
bt.insert(5);
bt.insert(3);
bt.insert(7);
bt.insert(1);
bt.insert(4);
bt.insert(6);
bt.insert(9);
bt.printTree();
//bt.delete(3);
//bt.delete(4);
//bt.delete(6);
bt.delete(9);
//bt.delete(5);
bt.printTree();
}
}
I'm going to provide you the logic (that means you have to write the code yourself) of how to delete a node in a BST.
There are three cases.
Node to be deleted has both left and right child as null: Delete the node and make the parent point to null.
Node to be deleted has either left or right child (but not both) as null: Delete the node but make sure that the parent points to the valid child of the to-be-deleted node.
Node to be deleted has nether left child nor right child as null: In this case, you have to find the next greater element of the to-be-deleted node. This next greater element is the least element of the right subtree of the to-be-deleted node. Since this is the least element, it has at least one of its child as null. So swap the values of the to-be-deleted node with the next greater node. After you swap, delete this next greater node using points 1 and 2 (whichever is fitting to the situation). Now, why the next greater node and not any node. Because if you replace a node with its next greater node, the BST remains a BST. Try it out in an example and it will be clear.

Finding BST's height non-recursively?

This is a recursive method for finding the height, but i have a very large number of nodes in my binary search tree, and i want to find the height of the tree as well as assign the height to each individual sub-tree. So the recursive method throws stackoverflow exception, how do i do this non-recursively and without using stack?
private int FindHeight(TreeNode node)
{
if (node == null)
{
return -1;
}
else
{
node.Height = 1 + Math.Max(FindHeight(node.Left), FindHeight(node.Right));
return node.Height;
}
}
I believe i have to use post order traversal but without stack?
I was able to make this method, and it does return the correct height but it assigns each node with its depth not height.
public void FindHeight()
{
int maxHeight = 0;
Queue<TreeNode> Q = new Queue<TreeNode>();
TreeNode node;
Q.Enqueue(Root);
while (Q.Count != 0)
{
node = Q.Dequeue();
int nodeHeight = node.Height;
if (node.Left != null)
{
node.Left.Height = nodeHeight + 1;
Q.Enqueue(node.Left);
}
if (node.Right != null)
{
node.Right.Height = nodeHeight + 1;
Q.Enqueue(node.Right);
}
if (nodeHeight > maxHeight)
{
maxHeight = nodeHeight;
}
}
Console.WriteLine(maxHeight);
}