解析 function isPalindrome(str) { str = str.replace(/\s/g, '').toLowerCase(); let left = 0; let right = str.length 1; while (left < right) { if (str[left] !== str[right]) { return false; } left++; right; } return true; }...
Another method to check if a string is a palindrome or not is by using aforloop. Below are the steps to check if a string is a palindrome in JavaScript. functionpalindromeFn(string){conststringLength=string.length;for(leti=0;i<stringLength/2;i++){if(string[i]!==string[stringLength-1...
JavaScript Code:// Helper function to check if a given string is a palindrome function isPalindrome(str) { return str === str.split('').reverse().join(''); } // Function to find the longest palindrome in a given string function longest_palindrome(str) { let maxLength = 0; // ...
Executable in both browser and server which has Javascript engines like V8(chrome), SpiderMonkey(Firefox) etc. Syntax help STDIN Example varreadline =require('readline');varrl = readline.createInterface({input: process.stdin,output: process.stdout,terminal:false}); rl.on('line',function(line){...
1.第一轮遍历用快慢指针(快指针每次走两步,慢指针每次走一步)寻找中点 -> O(n) 2.反转后半段链表 -> O(n/2) 3.比较 -> O(n/2) 合起来时间还是O(n)。 你确定这是easy? 1/**2* Definition for singly-linked list.3* function ListNode(val) {4* this.val = val;5* this.next = null;...
Here, we are going to learn how to check whether a given number is palindrome or not in JavaScript.
In contrast, the time complexity of the original solution is O(n + n + n), which simplifies to O(n). This is because the function creates a new string and a new array of characters, both of which have a length of n, and then performs a reverse operation that takes O(n) time. ...
Solution in javascript:function palindrome(str) { if(str.length > 0) { var middle = parseInt(str.length/2); for(var i = 0, j = str.length -1; i <= middle; i++) { if(str[i] === str[j]) { j--; } else { return false } } } else { return false } return true}palind...
In the function, We are checking for whether a string is an empty string or not – if the string is an empty string then throwing an error. Then, we are converting string to uppercase to make comparison case insensitive. Then, running a loop from 0 to len/2, to compare the first ...
Noticed that 1. Palindrome center only existing in the first half of the string. 2. If the center is not a single character, they should be same letters. 1/**2* @param {string} s3* @return {string}4*/5varshortestPalindrome =function(s) {6varprefix = "";7varpos, head, tail;89...