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<h2 class="wrapper title">LCCI 01.01.判定字符是否唯一</h2>
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<span>Author：</span><span>flypython</span> | <span>Date: </span><span>2020-02-25</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="01-01-判定字符是否唯一"><a href="#01-01-判定字符是否唯一" class="headerlink" title="01.01.判定字符是否唯一"></a>01.01.判定字符是否唯一</h2><p>链接：<a href="https://leetcode-cn.com/problems/is-unique-lcci/" target="_blank" rel="noopener">https://leetcode-cn.com/problems/is-unique-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>实现一个算法，确定一个字符串 s 的所有字符是否全都不同。</p>
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<p>示例 1：</p>
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<p>输入: s = “leetcode”<br>输出: false<br>示例 2：</p>
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<p>输入: s = “abc”<br>输出: true</p>
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<p>限制：</p>
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<p>0 <= len(s) <= 100<br>如果你不使用额外的数据结构，会很加分。</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>首先，你得理解题意。不确定的东西，需要问或者尝试验证。</p>
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<p>比如：这里你可以问面试官字符串s是ASCII还是Unicode字符串，如果是编程平台你可以写测试来判断。</p>
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<p>在这里我们假定字符集为ASCII，那如果字符串长度大于128，那肯定是false，不过有限制条件len(s)<=100，那这个判断就不必要了。限制里面还要求不使用额外的数据结构，这就过滤掉了很多的方案。</p>
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<p>比如，使用set就一行代码解决</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 isUnique(self, astr: str) -> bool:</span><br><span class="line"> return len(astr)== len(set(astr))</span><br></pre></td></tr></table></figure>
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<p>我们很容易想到hash计数的方法，经过测试，字符串中只出现’a’到’z’，26个小写字母，那只需要一个bool数组，直接统计出现的频率，判断是否重复。</p>
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<p>考虑到不使用额外的数据结构，那我们可以尝试使用bitmap，把数组换成一个整型数，按照位上0，1来判断是否重复。</p>
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<p>开始我们算出字符离最开始的字符’a’的距离，然后1移动这个距离来表示这个数。后面用一个测试变量来测试p，如果存在则返回False，如果不存在则写入。</p>
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<p>具体代码：</p>
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<pre><code>def isUnique(self, astr: str) -> bool:
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t = 0
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for c in astr:
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if t & (p := 1 << (ord(c) - ord('a'))):
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return False
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t |= p
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return True</code></pre><p>在解答中使用了Python3.8的新特性：海象运算符，点击下面的链接，学习使用它。<br><a href="https://mp.weixin.qq.com/s/3xKdt-26guYHoFb3xP0muw" target="_blank" rel="noopener">https://mp.weixin.qq.com/s/3xKdt-26guYHoFb3xP0muw</a></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></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def isUnique(self, astr: str) -> bool:</span><br><span class="line"> t = 0</span><br><span class="line"> for c in astr:</span><br><span class="line"> if t & (p := 1 << (ord(c) - ord('a'))):</span><br><span class="line"> return False</span><br><span class="line"> t |= p</span><br><span class="line"> return True</span><br></pre></td></tr></table></figure>
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<p>另外还有三种使用数据结构的解法</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></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def isUnique(self, astr: str) -> bool:</span><br><span class="line"> return len(astr)== len(set(astr))</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><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></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def isUnique(self, astr: str) -> bool:</span><br><span class="line"> adict={}</span><br><span class="line"> for value in astr:</span><br><span class="line"> if value in adict.keys():</span><br><span class="line"> return False</span><br><span class="line"> else:</span><br><span class="line"> adict[value]=0</span><br><span class="line"> return True</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></pre></td><td class="code"><pre><span class="line">class Solution:</span><br><span class="line"> def isUnique(self, astr: str) -> bool:</span><br><span class="line"> l=[]</span><br><span class="line"> for i in astr:</span><br><span class="line"> if i in l:</span><br><span class="line"> return False</span><br><span class="line"> else:</span><br><span class="line"> l.append(i)</span><br><span class="line"> return True</span><br></pre></td></tr></table></figure>
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