[LeetCode] 125. Valid Palindrome

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mob604756fca9f3 2020-01-15 03:37:00

文章标签 leetcode java javascript palindrome string 文章分类 后端开发

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

有效的回文。题意很简单，验证给的input是否是一个有效的回文。思路也很简单，two pointer逼近。注意

只比较input里面的字母和数字
需要ignore大小写

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {string} s
 3  * @return {boolean}
 4  */
 5 var isPalindrome = function (s) {
 6     if (s === null || s.length === 0) return true;
 7     const alphanum = s.toLowerCase().replace(/[\W]/g, "");
 8     let left = 0;
 9     let right = alphanum.length - 1;
10     while (left < right) {
11         if (alphanum[left] !== alphanum[right]) {
12             return false;
13         }
14         left++;
15         right--;
16     }
17     return true;
18 };

Java实现

 1 class Solution {
 2     public boolean isPalindrome(String s) {
 3         // corner case
 4         if (s == null || s.length() == 0) {
 5             return true;
 6         }
 7 
 8         // normal case
 9         int left = 0;
10         int right = s.length() - 1;
11         while (left < right) {
12             while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {
13                 left++;
14             }
15             while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {
16                 right--;
17             }
18             if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {
19                 return false;
20             }
21             left++;
22             right--;
23         }
24         return true;
25     }
26 }