# 0125. 驗證回文串 ## 題目地址(125. 驗證回文串) <https://leetcode-cn.com/problems/valid-palindrome/description/> ## 題目描述 ``` <pre class="calibre18">``` 給定一個字符串，驗證它是否是回文串，只考慮字母和數字字符，可以忽略字母的大小寫。 說明：本題中，我們將空字符串定義為有效的回文串。 示例 1: 輸入: "A man, a plan, a canal: Panama" 輸出: true 示例 2: 輸入: "race a car" 輸出: false ``` ``` ## 前置知識 - 回文 - 雙指針 ## 公司 - 阿里 - 騰訊 - 百度 - 字節 - facebook - microsoft - uber - zenefits ## 思路 這是一道考察回文的題目，而且是最簡單的形式，即判斷一個字符串是否是回文。 針對這個問題，我們可以使用頭尾雙指針， - 如果兩個指針的元素不相同，則直接返回 false, - 如果兩個指針的元素相同，我們同時更新頭尾指針，循環。 直到頭尾指針相遇。 時間復雜度為 O(n). 拿“noon”這樣一個回文串來說，我們的判斷過程是這樣的：  拿“abaa”這樣一個不是回文的字符串來說，我們的判斷過程是這樣的：  ## 關鍵點解析 - 雙指針 ## 代碼 - 語言支持：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">// 只處理英文字符（題目忽略大小寫，我們前面全部轉化成了小寫， 因此這里我們只判斷小寫）和數字</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">""" 使用語言特性進行求解 """</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>] ``` ``` **復雜度分析** - 時間復雜度：O(N)O(N)O(N) - 空間復雜度：O(1)O(1)O(1) 大家對此有何看法，歡迎給我留言，我有時間都會一一查看回答。更多算法套路可以訪問我的 LeetCode 題解倉庫：<https://github.com/azl397985856/leetcode> 。 目前已經 37K star 啦。 大家也可以關注我的公眾號《力扣加加》帶你啃下算法這塊硬骨頭。