百度：实现一个函数，判断输入是不是回文字符串 #20

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1. sisterAn

   changed the title [-]Leetcode3：无重复字符的最长子串[/-] [+]百度：实现一个函数，判断输入是不是回文字符串[/+] on Apr 19, 2020
2. Aiyoweioo commented on Apr 20, 2020

   

   Aiyoweioo

   on Apr 20, 2020

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   方法一： 用首尾两个指针，向中间扫描字符串，如果两指针指向的字符都一样， 这字符串就是一个回文。时间复杂度：O(n)，空间复杂度：O(1)
   方法二：从中间开始，向两头扩展查看指向的字符是否一致。

   /** 判断是不是回文字符串
    * @param {string} str 
    * return Boolean
    */
   function IsPalindrome(str) {
     if (str === null  || str.length < 1) {
       return false
     }
     let left = 0, right = str.length - 1
     while(left < right) {
       if(str.charAt(left) !== str.charAt(right)) {
         return false
       }
       ++left;
       --right;
     }
     return true
   }

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3. sisterAn commented on Apr 20, 2020

   

   sisterAn

   on Apr 20, 2020

   OwnerAuthor

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   解法一：使用API

   function isPlalindrome(input) {
     if (typeof input !== 'string') return false;
     return input.split('').reverse().join('') === input;
   }

   解法二：不使用API

   function isPlalindrome(input) {
     if (typeof input !== 'string') return false;
     let i = 0, j = input.length - 1
     while(i < j) {
         if(input.charAt(i) !== input.charAt(j)) return false
         i ++
         j --
     }
     return true
   }

   👍React with 👍9Reacted by jmty, Miss Chen, joegb, laiyingzeng, junziwang, Jet , linwu0909, demo and uwiew
4. zero9527 commented on Apr 29, 2020

   

   zero9527

   on Apr 29, 2020

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   /**
    * 判断字符串是否是回文字符串
    * @param {*} str
    */
   function isReverseString(str) {
     if (typeof str !== 'string') return false;
     // 方法1：reverse
     // return str.split('').reverse().join('') === str;
   
     // 方法2
     const strArr = str.split('');
     let reverseStr = '';
     let len = strArr.length - 1;
     while (len >= 0) reverseStr += strArr[len--];
     return str === reverseStr;
   }
   
   console.log(isReverseString('asddsa')); // true
   console.log(isReverseString('asdDsa')); // false
   console.log(isReverseString('13dffd31')); // true
5. sisterAn

   added

   百度

   on May 5, 2020
6. qianlongo commented on Aug 15, 2020

   

   qianlongo

   on Aug 15, 2020

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   const isPalindrome = () => {
    const len = str.length
     const halfLen = len / 2
     
   
     for (let i = 0; i < halfLen; i++) {
       if (str[i] !== str[ len - i - 1 ]) {
         return false
       }
     }
   
     return true
   }
   
7. thinkingc commented on Apr 6, 2021

   

   thinkingc

   on Apr 6, 2021

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   双指针

   function isReverseString(str) {
     var half = str.toString().length / 2 | 0
     var l = 0
     var r = str.length - 1
   
     while (l != half) {
       if (str[l++] != str[r--]) return false
     }
     return true
   }
   
   var { log } = console
   log(isReverseString('12321'))
   log(isReverseString('123321'))
   log(isReverseString('234'))
8. AlexZhang11 commented on May 31, 2024

   

   AlexZhang11

   on May 31, 2024

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   function isPalindrome(str){
       let i = 0,k = str.length-1
       let flg = true
       while(i<k&&flg){
           if(str.charAt(i)!==str.charAt(k)){
               flg= false
           }
           i++
           k--
       }
       return flg
   }
   
   console.log(isPalindrome('aabbccdccbbaa'))
   console.log(isPalindrome('aabbccdvsccbbaa'))