162 lines
16 KiB
HTML
162 lines
16 KiB
HTML
<!DOCTYPE html>
|
||
<html lang="zh-CN">
|
||
<head>
|
||
<head><meta name="generator" content="Hexo 3.9.0">
|
||
<!-- Title -->
|
||
|
||
<meta charset="utf-8">
|
||
<meta name="applicable-device" content="pc,mobile">
|
||
<meta name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=3.0, viewport-fit=cover">
|
||
<meta http-equiv="X-UA-Compatible" content="IE=Edge,chrome=1">
|
||
<meta name="author" content="flypython">
|
||
<meta name="designer" content="flypython">
|
||
<meta name="keywords" content="LCCI 01.02.判定是否互为字符重排,FlyPython - 专业的Python学习社区,flypython, 飞蟒，飞蟒Python，Python入门，Python自动化，Python日报">
|
||
<meta property="og:title" content="LCCI 01.02.判定是否互为字符重排 | FlyPython - 专业的Python学习社区">
|
||
<meta property="og:site_name" content="http://www.flypython.com">
|
||
|
||
<meta property="og:type" content="article">
|
||
<meta property="og:url" content="http://www.flypython.com/article/python-leetcode-lcci0102/">
|
||
<meta property="og:image" content="http://www.flypython.com/images/leetcode-0102.png">
|
||
<meta property="og:description" content="LCCI 01.02.判定是否互为字符重排--LeetCode刷题计划">
|
||
<meta name="description" content="LCCI 01.02.判定是否互为字符重排--LeetCode刷题计划">
|
||
|
||
<meta name="rating" content="general">
|
||
<meta name="apple-mobile-web-app-capable" content="yes">
|
||
<meta name="apple-mobile-web-app-status-bar-style" content="black">
|
||
<meta name="format-detection" content="telephone=yes">
|
||
<meta name="mobile-web-app-capable" content="yes">
|
||
<meta name="robots" content="index, follow">
|
||
<link rel="icon" href="/images/favicon.ico">
|
||
<title>LCCI 01.02.判定是否互为字符重排 | FlyPython - 专业的Python学习社区</title>
|
||
<link rel="stylesheet" href="/css/f25.css">
|
||
<link rel="stylesheet" href="/css/highlight.css">
|
||
|
||
|
||
<!-- Global site tag (gtag.js) - Google Analytics -->
|
||
<script async src="https://www.googletagmanager.com/gtag/js?id=UA-147288599-1"></script>
|
||
<script>
|
||
window.dataLayer = window.dataLayer || [];
|
||
function gtag(){dataLayer.push(arguments);}
|
||
gtag('js', new Date());
|
||
|
||
gtag('config', 'UA-147288599-1');
|
||
</script>
|
||
|
||
</head>
|
||
</head>
|
||
<body>
|
||
<header class="wrapper header-wrapper">
|
||
<div class="container header-nav-wrapper">
|
||
<div class="logo"><a href="/" title="FlyPython - 专业的Python学习社区"><h1 class="title">FlyPython</h1></a></div>
|
||
<nav class="nav-wrapper">
|
||
|
||
<a href="https://flypython.com/python" title="飞蟒微课堂">飞蟒微课堂</a>
|
||
|
||
<a href="https://flypython.com/flypython_daily" title="Python日报">Python日报</a>
|
||
|
||
<a href="https://flypython.com/PyCon/" title="PyCon">PyCon</a>
|
||
|
||
<a href="https://github.com/flypythoncom" title="Github">Github</a>
|
||
|
||
<a href="/article/about" title="关于">关于</a>
|
||
|
||
</nav>
|
||
<span class="btn-menu" id="J_header_menu">
|
||
<div class="inner">
|
||
<span class="line line-01"></span>
|
||
<span class="line line-02"></span>
|
||
<span class="line line-03"></span>
|
||
</div>
|
||
</span>
|
||
<div class="wrapper mb-nav-wrapper" id="J_header_menu_list">
|
||
<nav class="wrapper mb-nav-container">
|
||
|
||
<a href="https://flypython.com/python" title="飞蟒微课堂">飞蟒微课堂</a>
|
||
|
||
<a href="https://flypython.com/flypython_daily" title="Python日报">Python日报</a>
|
||
|
||
<a href="https://flypython.com/PyCon/" title="PyCon">PyCon</a>
|
||
|
||
<a href="https://github.com/flypythoncom" title="Github">Github</a>
|
||
|
||
<a href="/article/about" title="关于">关于</a>
|
||
|
||
</nav>
|
||
</div>
|
||
</div>
|
||
</header>
|
||
<section class="body-wrapper">
|
||
<section class="wrapper post-banner">
|
||
<div class="container post-banner-container">
|
||
<h2 class="wrapper title">LCCI 01.02.判定是否互为字符重排</h2>
|
||
<div class="wrapper tips">
|
||
<span>Author：</span><span>flypython</span> | <span>Date: </span><span>2020-02-26</span> | <span>Category：</span><span><a href="/fly/飞蟒微课堂/" title="飞蟒微课堂">飞蟒微课堂</a><a href="/fly/飞蟒微课堂/LeetCode/" title="LeetCode">LeetCode</a></span>
|
||
</div>
|
||
</div>
|
||
</section>
|
||
<section class="wrapper main-wrapper">
|
||
<article class="sub-container post-content">
|
||
<h2 id="01-02-判定是否互为字符重排"><a href="#01-02-判定是否互为字符重排" class="headerlink" title="01.02.判定是否互为字符重排"></a>01.02.判定是否互为字符重排</h2><p>链接：<a href="https://leetcode-cn.com/problems/check-permutation-lcci/" target="_blank" rel="noopener">https://leetcode-cn.com/problems/check-permutation-lcci/</a><br>难度：简单</p>
|
||
<h4 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h4><blockquote>
|
||
<p>给定两个字符串 s1 和 s2，请编写一个程序，确定其中一个字符串的字符重新排列后，能否变成另一个字符串。</p>
|
||
<p>示例 1：</p>
|
||
<p>输入: s1 = “abc”, s2 = “bca”<br>输出: true<br>示例 2：</p>
|
||
<p>输入: s1 = “abc”, s2 = “bad”<br>输出: false</p>
|
||
<p>说明：<br>0 <= len(s1) <= 100<br>0 <= len(s2) <= 100</p>
|
||
</blockquote>
|
||
<h4 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h4><p>开始，我们还是先理解题意，确定是否区分大小写，是否忽略空白字符等。有了01.01题的hash计数0，1判断的经历，在这里很容易想到使用hash计数。</p>
|
||
<p>因为字符可以重复，需要对字符进行计数，这个时候我们会想到collections库中的Counter。于是有了下面两行代码解决了问题。</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">from collections import Counter</span><br><span class="line">return Counter(s1) == Counter(s2)</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>但是上面的方法严格来说算作弊，那我们手写吧。对每个字符进行计数操作并测试是否相等，不相等返回False，全部相等返回True。</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def CheckPermutation(self, s1: str, s2: str) -> bool:</span><br><span class="line"> if len(s1) != len(s2):</span><br><span class="line"> return False</span><br><span class="line"> </span><br><span class="line"> base = set(s1+s2)</span><br><span class="line"> for b in base:</span><br><span class="line"> if s1.count(b) != s2.count(b):</span><br><span class="line"> return False</span><br><span class="line"> return True</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>上面的代码，最容易出问题的是<code>base = set(s1+s2)</code>,需要以两个字符串集的唯一字符来进行计数，但这里引入了set。</p>
|
||
<p>我们测试了字符范围，发现是’a’到’z’，那么代码可以改成：</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def CheckPermutation(self, s1: str, s2: str) -> bool:</span><br><span class="line"> if len(s1) != len(s2):</span><br><span class="line"> return False</span><br><span class="line"> </span><br><span class="line"> base = [chr(i) for i in range(ord("a"),ord("z")+1)]</span><br><span class="line"> for b in base:</span><br><span class="line"> if s1.count(b) != s2.count(b):</span><br><span class="line"> return False</span><br><span class="line"> return True</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>再进一步，为了直观，我们可以引入string包，代码如下：</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def CheckPermutation(self, s1: str, s2: str) -> bool:</span><br><span class="line"> import string </span><br><span class="line"> if len(s1) != len(s2):</span><br><span class="line"> return False</span><br><span class="line"> </span><br><span class="line"> for b in string.ascii_lowercase:</span><br><span class="line"> if s1.count(b) != s2.count(b):</span><br><span class="line"> return False</span><br><span class="line"> return True</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>当然，在这里我们也可以通过排序并比较来解决，代码非常简单:</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def CheckPermutation(self, s1: str, s2: str) -> bool:</span><br><span class="line"> return sorted(s1) == sorted(s2)</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>空间复杂度O(nlogn)，相比hash计数复杂度要高一些，但是hash计数有空间复杂度。</p>
|
||
<h4 id="错误代码分析"><a href="#错误代码分析" class="headerlink" title="错误代码分析"></a>错误代码分析</h4><p>在题解中看到一份代码</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line">class Solution {</span><br><span class="line"> public boolean CheckPermutation(String s1, String s2) {</span><br><span class="line"> if(s1.length() != s2.length()) {</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> int s1Sum = 0, s2Sum = 0;</span><br><span class="line"> for(int i = 0; i < s1.length(); i++) {</span><br><span class="line"> s1Sum += s1.charAt(i);</span><br><span class="line"> s2Sum += s2.charAt(i);</span><br><span class="line"> }</span><br><span class="line"> return s1Sum == s2Sum;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>此解法主要是计算s1，s2字符和，比较字符和。这种解法的问题就是忽略了和相同，但数字并不同的情况，比如 3+7 == 2+8。</p>
|
||
<p>除了计算s1，s2的字符和外，计算相对位移和，加s1字符减s2字符等方法都有此问题。</p>
|
||
<h4 id="方案代码"><a href="#方案代码" class="headerlink" title="方案代码"></a>方案代码</h4><p>解法:</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def CheckPermutation(self, s1: str, s2: str) -> bool:</span><br><span class="line"> if len(s1) != len(s2):</span><br><span class="line"> return False</span><br><span class="line"> </span><br><span class="line"> base = set(s1+s2)</span><br><span class="line"> for b in base:</span><br><span class="line"> if s1.count(b) != s2.count(b):</span><br><span class="line"> return False</span><br><span class="line"> return True</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>另外含高级数据结构的三种解法</p>
|
||
<p>解法1:</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def CheckPermutation(self, s1: str, s2: str) -> bool:</span><br><span class="line"> from collections import Counter</span><br><span class="line"> return Counter(s1) == Counter(s2)</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>解法2:</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def CheckPermutation(self, s1: str, s2: str) -> bool:</span><br><span class="line"> return sorted(s1) == sorted(s2)</span><br></pre></td></tr></table></figure>
|
||
|
||
<p>解法3:</p>
|
||
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def CheckPermutation(self, s1: str, s2: str) -> bool:</span><br><span class="line"> import string</span><br><span class="line"> if len(s1) != len(s2):</span><br><span class="line"> return False</span><br><span class="line"> </span><br><span class="line"> for b in string.ascii_lowercase:</span><br><span class="line"> if s1.count(b) != s2.count(b):</span><br><span class="line"> return False</span><br><span class="line"> return True</span><br></pre></td></tr></table></figure>
|
||
|
||
|
||
</article>
|
||
<div class="sub-container gitalk-wrapper" id="gitalk-container"></div>
|
||
</section>
|
||
|
||
<div class="tips-top-wrapper">
|
||
<span class="tip-top-container" onclick="scrollToWindowTop()">
|
||
<span class="l-bar"></span>
|
||
<span class="r-bar"></span>
|
||
</span>
|
||
</div>
|
||
<footer class="wrapper footer-wrapper">
|
||
<div class="container"><span class="copyright">© 2020 FlyPython . All Rights Reserved.</span></div>
|
||
</footer>
|
||
</section>
|
||
<script src="/js/f25.js"></script>
|
||
|
||
</body>
|
||
</html>
|