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<h2 class="wrapper title">LCCI 01.02.判定是否互为字符重排</h2>
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<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>
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<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>
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<h4 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h4><blockquote>
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<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>
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<p>示例 1：</p>
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<p>输入: s1 = “abc”, s2 = “bca”<br>输出: true<br>示例 2：</p>
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<p>输入: s1 = “abc”, s2 = “bad”<br>输出: false</p>
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<p>说明：<br>0 <= len(s1) <= 100<br>0 <= len(s2) <= 100</p>
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</blockquote>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
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<p>再进一步，为了直观，我们可以引入string包，代码如下：</p>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
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<p>另外含高级数据结构的三种解法</p>
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<p>解法1:</p>
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<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>
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<p>解法2:</p>
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<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>
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<p>解法3:</p>
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<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>
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<script src="/js/f25.js"></script>
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<ul class="container list-container">
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||||
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||||
<li class="post-i-wrapper">
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||||
<a href="/article/python-leetcode-lcci0102/" class="wrapper post-i-link" title="LCCI 01.02.判定是否互为字符重排">
|
||||
<div class="wrapper cover">
|
||||
<img src="/images/leetcode-0102.png" alt="LCCI 01.02.判定是否互为字符重排">
|
||||
</div>
|
||||
<h3 class="wrapper title" title="LCCI 01.02.判定是否互为字符重排">LCCI 01.02.判定是否互为字符重排</h3>
|
||||
<p class="wrapper subtitle">
|
||||
LeetCode刷题计划
|
||||
</p>
|
||||
<div class="wrapper time">2020-02-26</div>
|
||||
</a>
|
||||
</li>
|
||||
|
||||
<li class="post-i-wrapper">
|
||||
<a href="/article/python-leetcode-lcci0101/" class="wrapper post-i-link" title="LCCI 01.01.判定字符是否唯一">
|
||||
<div class="wrapper cover">
|
||||
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||||
@@ -95,6 +95,19 @@
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||||
<div class="wrapper index-wrapper">
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||||
<ul class="container list-container">
|
||||
|
||||
<li class="post-i-wrapper">
|
||||
<a href="/article/python-leetcode-lcci0102/" class="wrapper post-i-link" title="LCCI 01.02.判定是否互为字符重排">
|
||||
<div class="wrapper cover">
|
||||
<img src="/images/leetcode-0102.png" alt="LCCI 01.02.判定是否互为字符重排">
|
||||
</div>
|
||||
<h3 class="wrapper title" title="LCCI 01.02.判定是否互为字符重排">LCCI 01.02.判定是否互为字符重排</h3>
|
||||
<p class="wrapper subtitle">
|
||||
LeetCode刷题计划
|
||||
</p>
|
||||
<div class="wrapper time">2020-02-26</div>
|
||||
</a>
|
||||
</li>
|
||||
|
||||
<li class="post-i-wrapper">
|
||||
<a href="/article/python-leetcode-lcci0101/" class="wrapper post-i-link" title="LCCI 01.01.判定字符是否唯一">
|
||||
<div class="wrapper cover">
|
||||
|
||||
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images/leetcode-0102.png
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13
index.html
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|
||||
<div class="wrapper index-wrapper">
|
||||
<ul class="container list-container">
|
||||
|
||||
<li class="post-i-wrapper">
|
||||
<a href="/article/python-leetcode-lcci0102/" class="wrapper post-i-link" title="LCCI 01.02.判定是否互为字符重排">
|
||||
<div class="wrapper cover">
|
||||
<img src="/images/leetcode-0102.png" alt="LCCI 01.02.判定是否互为字符重排">
|
||||
</div>
|
||||
<h3 class="wrapper title" title="LCCI 01.02.判定是否互为字符重排">LCCI 01.02.判定是否互为字符重排</h3>
|
||||
<p class="wrapper subtitle">
|
||||
LeetCode刷题计划
|
||||
</p>
|
||||
<div class="wrapper time">2020-02-26</div>
|
||||
</a>
|
||||
</li>
|
||||
|
||||
<li class="post-i-wrapper">
|
||||
<a href="/article/python-leetcode-lcci0101/" class="wrapper post-i-link" title="LCCI 01.01.判定字符是否唯一">
|
||||
<div class="wrapper cover">
|
||||
|
||||
Reference in New Issue
Block a user