I don't understand the +=+ and ++ used in calculations in this code. If I use += and single plus sign instead the calculations do not work.
document.getElementById("addbtn").addEventListener("click",addnumbers);
}
function addnumbers() {
var answer = "";
answer +=+
document.getElementById("inputOne").value + +
document.getElementById("inputTwo").value
alert(answer)
};
Related
I'm wanting to slice a range which I can do in Javascfript but am struggling in kotlin.
my current code is:
internal class blah {
fun longestPalindrome(s: String): String {
var longestP = ""
for (i in 0..s.length) {
for (j in 1..s.length) {
var subS = s.slice(i, j)
if (subS === subS.split("").reversed().joinToString("") && subS.length > longestP.length) {
longestP = subS
}
}
}
return longestP
}
and the error I get is:
Type mismatch.
Required:
IntRange
Found:
Int
Is there a way around this keeping most of the code I have?
As the error message says, slice wants an IntRange, not two Ints. So, pass it a range:
var subS = s.slice(i..j)
By the way, there are some bugs in your code:
You need to iterate up to the length minus 1 since the range starts at 0. But the easier way is to grab the indices range directly: for (i in s.indices)
I assume j should be i or bigger, not 1 or bigger, or you'll be checking some inverted Strings redundantly. It should look like for (j in i until s.length).
You need to use == instead of ===. The second operator is for referential equality, which will always be false for two computed Strings, even if they are identical.
I know this is probably just practice, but even with the above fixes, this code will fail if the String contains any multi-code-unit code points or any grapheme clusters. The proper way to do this would be by turning the String into a list of grapheme clusters and then performing the algorithm, but this is fairly complicated and should probably rely on some String processing code library.
class Solution {
fun longestPalindrome(s: String): String {
var longestPal = ""
for (i in 0 until s.length) {
for (j in i + 1..s.length) {
val substring = s.substring(i, j)
if (substring == substring.reversed() && substring.length > longestPal.length) {
longestPal = substring
}
}
}
return longestPal
}
}
This code is now functioning but unfortunately is not optimized enough to get through all test cases.
I'm trying a leetcode challenge and am struggling to pass the challenge due to the speed of my code:
class Solution {
fun longestPalindrome(s: String): String {
var longestPal = ""
var substring = ""
for (i in 0..s.length) {
for (j in i + 1..s.length) {
substring = s.substring(i, j)
if (substring == substring.reversed() && substring.length > longestPal.length) {
longestPal = substring
}
}
}
return longestPal
}
}
I'm a newb and not familiar with Big O notation.
I imagine if I could use just one loop I would be able to speed this code up significantly but am not sure how I would go about this.
(Not saying this is the best approach, but that is a start)
Palindromes can only be found between two same letters. So one idea is to go through the string once, and keep track of letter indexes. When you encounter a letter you already saw before, and the difference in indexes is longer than the current longest palindrome, you check for a palindrome:
fun longestPal(s: String): String {
val letters = mutableMapOf<Char, MutableList<Int>>()
var longest = ""
s.indices.forEach { index ->
val indicesForCurrentChar = letters[s[index]] ?: mutableListOf()
for (i in indicesForCurrentChar) {
if ((index - i) < longest.length) break // (1) won't be the longest anyway
val sub = s.substring(i, index + 1)
if (sub == sub.reversed()) longest = sub
}
indicesForCurrentChar.add(index)
letters[s[index]] = indicesForCurrentChar
}
return longest
}
What is costly here is the palindrome check itself (sub == sub.reversed). But the check in (1) should contain it (think of a string which is the repetition of the same letter).
I would be curious to know what other suggest online.
Your code runs in O(n^3) time, a loop within a loop within a loop, because that reversed() call iterates up to the size of the input string. You can look up Manacher's algorithm for an explanation of how to do it in linear time (O(n)), no nested iteration at all.
I've read somewhere that a variable should be entered into the code if it is reused. But when I write my code for logic transparency, I sometimes create intermediate variables (with names reflecting what they contain) which are used only once.
How incorrect is this concept?
PS:
I want to do it right.
It is important to note that most of the time clarity takes precedence over re-usability or brevity. This is one of the basic principles of clean code. Most modern compilers optimize code anyway so creating new variables need not be a concern at all.
It is perfectly fine to create a new variable if it would add clarity to your code. Make sure to give it a meaningful name. Consider the following function:
public static boolean isLeapYear(final int yyyy) {
if ((yyyy % 4) != 0) {
return false;
}
else if ((yyyy % 400) == 0) {
return true;
}
else if ((yyyy % 100) == 0) {
return false;
}
else {
return true;
}
}
Even though the boolean expressions are used only once, they may confuse the reader of the code. We can rewrite it as follows
public static boolean isLeapYear(int year) {
boolean fourth = year % 4 == 0;
boolean hundredth = year % 100 == 0;
boolean fourHundredth = year % 400 == 0;
return fourth && (!hundredth || fourHundredth);
}
These boolean variables add much more clarity to the code.
This example is from the Clean Code book by Robert C. Martin.
How can I make it in kotlin using for loop?
for (double i = 0; i < 10.0; i += 0.25) {
System.out.println("value is:" + i);
}
You should use the Intellij plugin for converting Java code for Kotlin. It's pretty neat (unless you have complex code using lambdas) This is what it converts to for your given question:
var i = 0.0
while (i < 10.0) {
println("value is:" + i)
i += 0.25
}
Here is the kotlin code equivalent to for loop.
var i = 0.0
while (i < 10.0)
{
println("value is:" + i)
i += 1.0
}
Kotlin for loop only support iterate the arrays.Please refer https://kotlinlang.org/docs/reference/control-flow.html
It's achievable in different way
var length:Double = 10.0
var increment:Double = 0.25
for (index in Array((length/increment).toInt(), { i -> (i * increment) }))
println(index)
I'm not sure if this syntax is new, but natural numbers may be used to iterate values like so.
(0..(10*4)).map {
it / 4.0 as Double
}.forEach {
println(it)
}
I'm avoiding iteration on IEEE 754 floating points, as that could potentially cause errors. It may be fine in this case, but the same may not be true depending on environment, context, and language. It's best to avoid using floating point except in cases where the numbers are expected to have arbitrary precision, i.e. uses real numbers or continuity.
This syntax may be used within a for loop as well if you don't need to store the result. However, for loops should be avoided for the most part if you want immutability (which you should).
for (i in 0..10) {
println("Project Loom is a lie! $i")
}
I want to keep white space when I call text attribute of token, is there any way to do it?
Here is the situation:
We have the following code
IF L > 40 THEN;
ELSE
IF A = 20 THEN
PUT "HELLO";
In this case, I want to transform it into:
if (!(L>40){
if (A=20)
put "hello";
}
The rule in Antlr is that:
stmt_if_block: IF expr
THEN x=stmt
(ELSE y=stmt)?
{
if ($x.text.equalsIgnoreCase(";"))
{
WriteLn("if(!(" + $expr.text +")){");
WriteLn($stmt.text);
Writeln("}");
}
}
But the result looks like:
if(!(L>40))
{
ifA=20put"hello";
}
The reason is that the white space in $stmt was removed. I was wondering if there is anyway to keep these white space
Thank you so much
Update: If I add
SPACE: [ ] -> channel(HIDDEN);
The space will be preserved, and the result would look like below, many spaces between tokens:
IF SUBSTR(WNAME3,M-1,1) = ')' THEN M = L; ELSE M = L - 1;
This is the C# extension method I use for exactly this purpose:
public static string GetFullText(this ParserRuleContext context)
{
if (context.Start == null || context.Stop == null || context.Start.StartIndex < 0 || context.Stop.StopIndex < 0)
return context.GetText(); // Fallback
return context.Start.InputStream.GetText(Interval.Of(context.Start.StartIndex, context.Stop.StopIndex));
}
Since you're using java, you'll have to translate it, but it should be straightforward - the API is the same.
Explanation: Get the first token, get the last token, and get the text from the input stream between the first char of the first token and the last char of the last token.
#Lucas solution, but in java in case you have troubles in translating:
private String getFullText(ParserRuleContext context) {
if (context.start == null || context.stop == null || context.start.getStartIndex() < 0 || context.stop.getStopIndex() < 0)
return context.getText();
return context.start.getInputStream().getText(Interval.of(context.start.getStartIndex(), context.stop.getStopIndex()));
}
Looks like InputStream is not always updated after removeLastChild/addChild operations. This solution helped me for one grammar, but it doesn't work for another.
Works for this grammar.
Doesn't work for modern groovy grammar (for some reason inputStream.getText contains old text).
I am trying to implement function name replacement like this:
enterPostfixExpression(ctx: PostfixExpressionContext) {
// Get identifierContext from ctx
...
const token = CommonTokenFactory.DEFAULT.createSimple(GroovyParser.Identifier, 'someNewFnName');
const node = new TerminalNode(token);
identifierContext.removeLastChild();
identifierContext.addChild(node);
UPD: I used visitor pattern for the first implementation