问题
I wonder how to write palindrome in javascript, where I input different words and program shows if word is palindrome or not. For example word noon is palindrome, while bad is not.
Thank you in advance.
回答1:
function palindrome(str) {
var len = str.length;
var mid = Math.floor(len/2);
for ( var i = 0; i < mid; i++ ) {
if (str[i] !== str[len - 1 - i]) {
return false;
}
}
return true;
}
palindrome will return if specified word is palindrome, based on boolean value (true/false)
UPDATE:
I opened bounty on this question due to performance and I've done research and here are the results:
If we are dealing with very large amount of data like
var abc = "asdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfd";
for ( var i = 0; i < 10; i++ ) {
abc += abc; // making string even more larger
}
function reverse(s) { // using this method for second half of string to be embedded
return s.split("").reverse().join("");
}
abc += reverse(abc); // adding second half string to make string true palindrome
In this example palindrome is True, just to note
Posted palindrome function gives us time from 180 to 210 Milliseconds (in current example), and the function posted below
with string == string.split('').reverse().join('') method gives us 980 to 1010 Milliseconds.
Machine Details:
System: Ubuntu 13.10 OS Type: 32 Bit RAM: 2 Gb CPU: 3.4 Ghz*2 Browser: Firefox 27.0.1
回答2:
Try this:
var isPalindrome = function (string) {
if (string == string.split('').reverse().join('')) {
alert(string + ' is palindrome.');
}
else {
alert(string + ' is not palindrome.');
}
}
document.getElementById('form_id').onsubmit = function() {
isPalindrome(document.getElementById('your_input').value);
}
So this script alerts the result, is it palindrome or not. You need to change the your_id with your input id and form_id with your form id to get this work.
Demo!
回答3:
Use something like this
function isPalindrome(s) {
return s == s.split("").reverse().join("") ? true : false;
}
alert(isPalindrome("noon"));
alternatively the above code can be optimized as [updated after rightfold's comment]
function isPalindrome(s) {
return s == s.split("").reverse().join("");
}
alert(isPalindrome("malayalam"));
alert(isPalindrome("english"));
回答4:
Look at this:
function isPalindrome(word){
if(word==null || word.length==0){
// up to you if you want true or false here, don't comment saying you
// would put true, I put this check here because of
// the following i < Math.ceil(word.length/2) && i< word.length
return false;
}
var lastIndex=Math.ceil(word.length/2);
for (var i = 0; i < lastIndex && i< word.length; i++) {
if (word[i] != word[word.length-1-i]) {
return false;
}
}
return true;
}
Edit: now half operation of comparison are performed since I iterate only up to half word to compare it with the last part of the word. Faster for large data!!!
Since the string is an array of char no need to use charAt functions!!!
Reference: http://wiki.answers.com/Q/Javascript_code_for_palindrome
回答5:
Faster Way:
-Compute half the way in loop.
-Store length of the word in a variable instead of calculating every time.
EDIT: Store word length/2 in a temporary variable as not to calculate every time in the loop as pointed out by (mvw) .
function isPalindrome(word){
var i,wLength = word.length-1,wLengthToCompare = wLength/2;
for (i = 0; i <= wLengthToCompare ; i++) {
if (word.charAt(i) != word.charAt(wLength-i)) {
return false;
}
}
return true;
}
回答6:
Let us start from the recursive definition of a palindrome:
- The empty string '' is a palindrome
- The string consisting of the character c, thus 'c', is a palindrome
- If the string s is a palindrome, then the string 'c' + s + 'c' for some character c is a palindrome
This definition can be coded straight into JavaScript:
function isPalindrome(s) {
var len = s.length;
// definition clauses 1. and 2.
if (len < 2) {
return true;
}
// note: len >= 2
// definition clause 3.
if (s[0] != s[len - 1]) {
return false;
}
// note: string is of form s = 'a' + t + 'a'
// note: s.length >= 2 implies t.length >= 0
var t = s.substr(1, len - 2);
return isPalindrome(t);
}
Here is some additional test code for MongoDB's mongo JavaScript shell, in a web browser with debugger replace print() with console.log()
function test(s) {
print('isPalindrome(' + s + '): ' + isPalindrome(s));
}
test('');
test('a');
test('ab');
test('aa');
test('aab');
test('aba');
test('aaa');
test('abaa');
test('neilarmstronggnortsmralien');
test('neilarmstrongxgnortsmralien');
test('neilarmstrongxsortsmralien');
I got this output:
$ mongo palindrome.js
MongoDB shell version: 2.4.8
connecting to: test
isPalindrome(): true
isPalindrome(a): true
isPalindrome(ab): false
isPalindrome(aa): true
isPalindrome(aab): false
isPalindrome(aba): true
isPalindrome(aaa): true
isPalindrome(abaa): false
isPalindrome(neilarmstronggnortsmralien): true
isPalindrome(neilarmstrongxgnortsmralien): true
isPalindrome(neilarmstrongxsortsmralien): false
An iterative solution is:
function isPalindrome(s) {
var len = s.length;
if (len < 2) {
return true;
}
var i = 0;
var j = len - 1;
while (i < j) {
if (s[i] != s[j]) {
return false;
}
i += 1;
j -= 1;
}
return true;
}
回答7:
Taking a stab at this. Kind of hard to measure performance, though.
function palin(word) {
var i = 0,
len = word.length - 1,
max = word.length / 2 | 0;
while (i < max) {
if (word.charCodeAt(i) !== word.charCodeAt(len - i)) {
return false;
}
i += 1;
}
return true;
}
My thinking is to use charCodeAt() instead charAt() with the hope that allocating a Number instead of a String will have better perf because Strings are variable length and might be more complex to allocate. Also, only iterating halfway through (as noted by sai) because that's all that's required. Also, if the length is odd (ex: 'aba'), the middle character is always ok.
回答8:
The most important thing to do when solving a Technical Test is Don't use shortcut methods -- they want to see how you think algorithmically! Not your use of methods.
Here is one that I came up with (45 minutes after I blew the test). There are a couple optimizations to make though. When writing any algorithm, its best to assume false and alter the logic if its looking to be true.
isPalindrome():
Basically, to make this run in O(N) (linear) complexity you want to have 2 iterators whose vectors point towards each other. Meaning, one iterator that starts at the beginning and one that starts at the end, each traveling inward. You could have the iterators traverse the whole array and use a condition to break/return once they meet in the middle, but it may save some work to only give each iterator a half-length by default.
for loops seem to force the use of more checks, so I used while loops - which I'm less comfortable with.
Here's the code:
/**
* TODO: If func counts out, let it return 0
* * Assume !isPalindrome (invert logic)
*/
function isPalindrome(S){
var s = S
, len = s.length
, mid = len/2;
, i = 0, j = len-1;
while(i<mid){
var l = s.charAt(i);
while(j>=mid){
var r = s.charAt(j);
if(l === r){
console.log('@while *', i, l, '...', j, r);
--j;
break;
}
console.log('@while !', i, l, '...', j, r);
return 0;
}
++i;
}
return 1;
}
var nooe = solution('neveroddoreven'); // even char length
var kayak = solution('kayak'); // odd char length
var kayaks = solution('kayaks');
console.log('@isPalindrome', nooe, kayak, kayaks);
Notice that if the loops count out, it returns true. All the logic should be inverted so that it by default returns false. I also used one short cut method String.prototype.charAt(n), but I felt OK with this as every language natively supports this method.
回答9:
Best Way to check string is palindrome with more criteria like case and special characters...
function checkPalindrom(str) {
var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
return str == str.split('').reverse().join('');
}
You can test it with following words and strings and gives you more specific result.
1. bob
2. Doc, note, I dissent. A fast never prevents a fatness. I diet on cod
For strings it ignores special characters and convert string to lower case.
回答10:
function palindrome(str) {
var lenMinusOne = str.length - 1;
var halfLen = Math.floor(str.length / 2);
for (var i = 0; i < halfLen; ++i) {
if (str[i] != str[lenMinusOne - i]) {
return false;
}
}
return true;
}
Optimized for half string parsing and for constant value variables.
回答11:
I think following function with time complexity of o(log n) will be better.
function palindrom(s){
s = s.toString();
var f = true; l = s.length/2, len = s.length -1;
for(var i=0; i < l; i++){
if(s[i] != s[len - i]){
f = false;
break;
}
}
return f;
}
console.log(palindrom(12321));
回答12:
Here's another way of doing it:
function isPalin(str) {
str = str.replace(/\W/g,'').toLowerCase();
return(str==str.split('').reverse().join(''));
}
回答13:
25x faster + recursive + non-branching + terse
function isPalindrome(s,i) {
return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}
See my complete explanation here.
回答14:
function palindrome(str) {
var re = /[^A-Za-z0-9]/g;
str = str.toLowerCase().replace(re, '');
var len = str.length;
for (var i = 0; i < len/2; i++) {
if (str[i] !== str[len - 1 - i]) {
return false;
}
}
return true;
}
回答15:
Or you could do it like this.
var palindrome = word => word == word.split('').reverse().join('')
回答16:
How about this one?
function pall (word) {
var lowerCWord = word.toLowerCase();
var rev = lowerCWord.split('').reverse().join('');
return rev.startsWith(lowerCWord);
}
pall('Madam');
回答17:
This function will remove all non-alphanumeric characters (punctuation, spaces, and symbols) and turn everything lower case in order to check for palindromes.
function palindrome(str){
var re = /[^A-Za-z0-9]/g;
str = str.toLowerCase().replace(re, '');
return str == str.split('').reverse().join('') ? true : false;
}
回答18:
Here's a one-liner without using String.reverse,
const isPal = str => Array
.apply(null, new Array(strLen = str.length))
.reduce((acc, s, i) => acc + str[strLen - (i + 1)], '') === str;
回答19:
str1 is the original string with deleted non-alphanumeric characters and spaces and str2 is the original string reversed.
function palindrome(str) {
var str1 = str.toLowerCase().replace(/\s/g, '').replace(
/[^a-zA-Z 0-9]/gi, "");
var str2 = str.toLowerCase().replace(/\s/g, '').replace(
/[^a-zA-Z 0-9]/gi, "").split("").reverse().join("");
if (str1 === str2) {
return true;
}
return false;
}
palindrome("almostomla");
回答20:
Note: This is case sensitive
function palindrome(word)
{
for(var i=0;i<word.length/2;i++)
if(word.charAt(i)!=word.charAt(word.length-(i+1)))
return word+" is Not a Palindrome";
return word+" is Palindrome";
}
Here is the fiddle: http://jsfiddle.net/eJx4v/
回答21:
I am not sure how this JSPerf check the code performance. I just tried to reverse the string & check the values. Please comment about the Pros & Cons of this method.
function palindrome(str) {
var re = str.split(''),
reArr = re.slice(0).reverse();
for (a = 0; a < re.length; a++) {
if (re[a] == reArr[a]) {
return false;
} else {
return true;
}
}
}
JS Perf test
回答22:
function palindrome(str){
for (var i = 0; i <= str.length; i++){
if (str[i] !== str[str.length - 1 - i]) {
return "The string is not a palindrome";
}
}
return "The string IS a palindrome"
}
palindrome("abcdcba"); //"The string IS a palindrome"
palindrome("abcdcb"); //"The string is not a palindrome";
If you console.log this line: console.log(str[i] + " and " + str[str.length - 1 - i]), before the if statement, you'll see what (str[str.length - 1 - i]) is. I think this is the most confusing part but you'll get it easily when you check it out on your console.
回答23:
All these loops! How about some functional goodness :) May run in to tail call issues on old/current js engines, solved in ES6
function isPalendrome(str){
var valid = false;
if(str.length < 2 ) return true;
function even(i,ii){
str[i]===str[ii] ? ((i+1 !== ii) ? even(i+1,ii-1) : valid = true) : null
}
function odd(i, ii){
str[i]===str[ii] ? ((i !== ii) ? odd(i+1,ii-1) : valid = true) : null
}
if(str.length % 2){
return odd(0,str.length-1),valid;
}else{
return even(0,str.length-1),valid;
}
}
To test your call stack run this code, you will be able to parse strings double the call stack size
function checkStackSize(){
var runs = 70000;
var max_process = 1;
var max = 0;
function recurse_me() {
max_process++;
if(max_process === runs) return;
max = max_process;
try {
recurse_me()
} catch(e) {
max = max_process;
}
}
recurse_me()
console.log(max);
}
Due to the symmetrical nature of the problem you could chunk the string from the outside in and process the chunks that are within call stack limits.
by that I mean if the palindromes length is 1000. You could join 0-250 and 750-1000 and join 250-499 with 500-749. You can then pass each chunk in to the function. The advantage to this is you could run the process in parallel using web workers or threads for very large data sets.
回答24:
Below code tells how to get a string from textBox and tell you whether it is a palindrome are not & displays your answer in another textbox
<html>
<head>
<meta charset="UTF-8"/>
<link rel="stylesheet" href=""/>
</head>
<body>
<h1>1234</h1>
<div id="demo">Example</div>
<a accessKey="x" href="http://www.google.com" id="com" >GooGle</a>
<h1 id="tar">"This is a Example Text..."</h1>
Number1 : <input type="text" name="txtname" id="numb"/>
Number2 : <input type="text" name="txtname2" id="numb2"/>
Number2 : <input type="text" name="txtname3" id="numb3" />
<button type="submit" id="sum" onclick="myfun()" >count</button>
<button type="button" id="so2" onclick="div()" >counnt</button><br/><br/>
<ol>
<li>water</li>
<li>Mazaa</li>
</ol><br/><br/>
<button onclick="myfun()">TryMe</button>
<script>
function myfun(){
var pass = document.getElementById("numb").value;
var rev = pass.split("").reverse().join("");
var text = document.getElementById("numb3");
text.value = rev;
if(pass === rev){
alert(pass + " is a Palindrome");
}else{
alert(pass + " is Not a Palindrome")
}
}
</script>
</body>
</html>
回答25:
You could also do something like this :
function isPalindrome(str) {
var newStr = '';
for(var i = str.length - 1; i >=0; i--) {
newStr += str[i];
}
if(newStr == str) {
return true;
return newStr;
} else {
return false;
return newStr;
}
}
回答26:
ES6 way of doing it. Notice that I take advantage of the array method reduceRight to reverse a string (you can use array methods on strings if you give the string as context, as low level - strings are arrays of chars). No it is not as performant as other solutions, but didn't see any answer that came it it using es6 or higher order functions so figured I'd throw this one out there.
const palindrome = str => {
const middle = str.length/2;
const left = str.slice(0, middle)
const right = Array.prototype.reduceRight.call(str.slice(Math.round(middle)), (str, char) => str + char, '')
return left === right;
}
回答27:
To avoid errors with special characters use this function below
function palindrome(str){
var removeChar = str.replace(/[^A-Z0-9]/ig, "").toLowerCase();
var checkPalindrome = removeChar.split('').reverse().join('');
if(removeChar === checkPalindrome){
return true;
}else{
return false;
}
}
回答28:
The code is concise quick fast and understandable.
TL;DR
Explanation :
Here isPalindrome function accepts a str parameter which is typeof string.
- If the length of the str param is less than or equal to one it simply returns "false".
If the above case is false then it moves on to the second if statement and checks that if the character at 0 position of the string is same as character at the last place. It does an inequality test between the both.
str.charAt(0) // gives us the value of character in string at position 0 str.slice(-1) // gives us the value of last character in the string.
If the inequality result is true then it goes ahead and returns false.
- If result from the previous statement is false then it recursively calls the
isPalindrome(str)function over and over again until the final result.
function isPalindrome(str){
if (str.length <= 1) return true;
if (str.charAt(0) != str.slice(-1)) return false;
return isPalindrome(str.substring(1,str.length-1));
};
document.getElementById('submit').addEventListener('click',function(){
var str = prompt('whats the string?');
alert(isPalindrome(str))
});
document.getElementById('ispdrm').onsubmit = function(){alert(isPalindrome(document.getElementById('inputTxt').value));
}<!DOCTYPE html>
<html>
<body>
<form id='ispdrm'><input type="text" id="inputTxt"></form>
<button id="submit">Click me</button>
</body>
</html>回答29:
function palindrome(s) {
var re = /[\W_]/g;
var lowRegStr = s.toLowerCase().replace(re, '');
var reverseStr = lowRegStr.split('').reverse().join('');
return reverseStr === lowRegStr;
}
回答30:
For better performance you can also use this one
function palindrome(str) {
str = str.split("");
var i = str.length;
var check = "Yes"
if (i > 1) {
for (var j = 0; j < i / 2; j++) {
if (str[j] != str[i - 1 - j]) {
check = "NO";
break;
}
}
console.log(check);
} else {
console.log("YES");
}
}
来源：https://stackoverflow.com/questions/22111507/how-to-write-palindrome-in-javascript