# 0125. 验证回文串 ## 题目地址(125. 验证回文串) <https://leetcode-cn.com/problems/valid-palindrome/description/> ## 题目描述 ``` <pre class="calibre18">``` 给定一个字符串，验证它是否是回文串，只考虑字母和数字字符，可以忽略字母的大小写。 说明：本题中，我们将空字符串定义为有效的回文串。 示例 1: 输入: "A man, a plan, a canal: Panama" 输出: true 示例 2: 输入: "race a car" 输出: false ``` ``` ## 前置知识 - 回文 - 双指针 ## 公司 - 阿里 - 腾讯 - 百度 - 字节 - facebook - microsoft - uber - zenefits ## 思路 这是一道考察回文的题目，而且是最简单的形式，即判断一个字符串是否是回文。 针对这个问题，我们可以使用头尾双指针， - 如果两个指针的元素不相同，则直接返回 false, - 如果两个指针的元素相同，我们同时更新头尾指针，循环。 直到头尾指针相遇。 时间复杂度为 O(n). 拿“noon”这样一个回文串来说，我们的判断过程是这样的：  拿“abaa”这样一个不是回文的字符串来说，我们的判断过程是这样的：  ## 关键点解析 - 双指针 ## 代码 - 语言支持：JS，C++，Python JavaScript Code: ``` <pre class="calibre18">``` <span class="hljs-title">/* * @lc app=leetcode id=125 lang=javascript * * [125] Valid Palindrome */</span> <span class="hljs-title">// 只处理英文字符（题目忽略大小写，我们前面全部转化成了小写， 因此这里我们只判断小写）和数字</span> <span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">isValid</span>(<span class="hljs-params">c</span>) </span>{ <span class="hljs-keyword">const</span> charCode = c.charCodeAt(<span class="hljs-params">0</span>); <span class="hljs-keyword">const</span> isDigit = charCode >= <span class="hljs-string">"0"</span>.charCodeAt(<span class="hljs-params">0</span>) && charCode <= <span class="hljs-string">"9"</span>.charCodeAt(<span class="hljs-params">0</span>); <span class="hljs-keyword">const</span> isChar = charCode >= <span class="hljs-string">"a"</span>.charCodeAt(<span class="hljs-params">0</span>) && charCode <= <span class="hljs-string">"z"</span>.charCodeAt(<span class="hljs-params">0</span>); <span class="hljs-keyword">return</span> isDigit || isChar; } <span class="hljs-title">/** * @param {string} s * @return {boolean} */</span> <span class="hljs-keyword">var</span> isPalindrome = <span class="hljs-function"><span class="hljs-keyword">function</span> (<span class="hljs-params">s</span>) </span>{ s = s.toLowerCase(); <span class="hljs-keyword">let</span> left = <span class="hljs-params">0</span>; <span class="hljs-keyword">let</span> right = s.length - <span class="hljs-params">1</span>; <span class="hljs-keyword">while</span> (left < right) { <span class="hljs-keyword">if</span> (!isValid(s[left])) { left++; <span class="hljs-keyword">continue</span>; } <span class="hljs-keyword">if</span> (!isValid(s[right])) { right--; <span class="hljs-keyword">continue</span>; } <span class="hljs-keyword">if</span> (s[left] === s[right]) { left++; right--; } <span class="hljs-keyword">else</span> { <span class="hljs-keyword">break</span>; } } <span class="hljs-keyword">return</span> right <= left; }; ``` ``` C++ Code: ``` <pre class="calibre18">``` <span class="hljs-keyword">class</span> Solution { <span class="hljs-keyword">public</span>: <span class="hljs-function"><span class="hljs-keyword">bool</span> <span class="hljs-title">isPalindrome</span><span class="hljs-params">(<span class="hljs-params">string</span> s)</span> </span>{ <span class="hljs-keyword">if</span> (s.empty()) <span class="hljs-keyword">return</span> <span class="hljs-params">true</span>; <span class="hljs-keyword">const</span> <span class="hljs-keyword">char</span>* s1 = s.c_str(); <span class="hljs-keyword">const</span> <span class="hljs-keyword">char</span>* e = s1 + s.length() - <span class="hljs-params">1</span>; <span class="hljs-keyword">while</span> (e > s1) { <span class="hljs-keyword">if</span> (!<span class="hljs-params">isalnum</span>(*s1)) {++s1; <span class="hljs-keyword">continue</span>;} <span class="hljs-keyword">if</span> (!<span class="hljs-params">isalnum</span>(*e)) {--e; <span class="hljs-keyword">continue</span>;} <span class="hljs-keyword">if</span> (<span class="hljs-params">tolower</span>(*s1) != <span class="hljs-params">tolower</span>(*e)) <span class="hljs-keyword">return</span> <span class="hljs-params">false</span>; <span class="hljs-keyword">else</span> {--e; ++s1;} } <span class="hljs-keyword">return</span> <span class="hljs-params">true</span>; } }; ``` ``` Python Code: ``` <pre class="calibre18">``` <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">Solution</span>:</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">isPalindrome</span><span class="hljs-params">(self, s: str)</span> -> bool:</span> left, right = <span class="hljs-params">0</span>, len(s) - <span class="hljs-params">1</span> <span class="hljs-keyword">while</span> left < right: <span class="hljs-keyword">if</span> <span class="hljs-keyword">not</span> s[left].isalnum(): left += <span class="hljs-params">1</span> <span class="hljs-keyword">continue</span> <span class="hljs-keyword">if</span> <span class="hljs-keyword">not</span> s[right].isalnum(): right -= <span class="hljs-params">1</span> <span class="hljs-keyword">continue</span> <span class="hljs-keyword">if</span> s[left].lower() == s[right].lower(): left += <span class="hljs-params">1</span> right -= <span class="hljs-params">1</span> <span class="hljs-keyword">else</span>: <span class="hljs-keyword">break</span> <span class="hljs-keyword">return</span> right <= left <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">isPalindrome2</span><span class="hljs-params">(self, s: str)</span> -> bool:</span> <span class="hljs-string">""" 使用语言特性进行求解 """</span> s = <span class="hljs-string">''</span>.join(i <span class="hljs-keyword">for</span> i <span class="hljs-keyword">in</span> s <span class="hljs-keyword">if</span> i.isalnum()).lower() <span class="hljs-keyword">return</span> s == s[::<span class="hljs-params">-1</span>] ``` ``` **复杂度分析** - 时间复杂度：O(N)O(N)O(N) - 空间复杂度：O(1)O(1)O(1) 大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库：<https://github.com/azl397985856/leetcode> 。 目前已经 37K star 啦。 大家也可以关注我的公众号《力扣加加》带你啃下算法这块硬骨头。