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</head><body><header><a href="/" class="logo-link"><img src="/logo.png"></a><ul class="nav nav-list"><li class="nav-list-item"><a href="/" target="_self" class="nav-list-link">ALL</a></li><li class="nav-list-item"><a href="/categories/leetcode/" target="_self" class="nav-list-link">LEETCODE</a></li><li class="nav-list-item"><a href="/atom.xml" target="_self" class="nav-list-link">RSS</a></li></ul></header><section class="container"><div class="post"><article class="post-block"><h1 class="post-title">leetcode-1071</h1><div class="post-meta"><div class="post-time">2020年3月30日</div></div><div class="post-content"><h3 id="1071-字符串的最大公因子"><a href="#1071-字符串的最大公因子" class="headerlink" title="1071. 字符串的最大公因子"></a>1071. 字符串的最大公因子</h3><p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/greatest-common-divisor-of-strings/">题目</a></p>
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<p>如果存在这样字符串，那它最大的长度就是这两个字符串长度的最大公约数。</p>
<figure class="highlight python"><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><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line">  <span class="function"><span class="keyword">def</span> <span class="title">gcdOfStrings</span>(<span class="params">self, str1: <span class="built_in">str</span>, str2: <span class="built_in">str</span></span>) -&gt; <span class="built_in">str</span>:</span></span><br><span class="line">    <span class="keyword">if</span> str1[<span class="number">0</span>] != str2[<span class="number">0</span>]:</span><br><span class="line">      <span class="keyword">return</span> <span class="string">&#x27;&#x27;</span></span><br><span class="line"></span><br><span class="line">    a = <span class="built_in">len</span>(str1)</span><br><span class="line">    b = <span class="built_in">len</span>(str2)</span><br><span class="line">    <span class="built_in">print</span>(a, b)</span><br><span class="line">    <span class="keyword">if</span> a &lt; b:</span><br><span class="line">      str1, str2 = str2, str1</span><br><span class="line">      a = <span class="built_in">len</span>(str1)</span><br><span class="line">      b = <span class="built_in">len</span>(str2)</span><br><span class="line"></span><br><span class="line">    <span class="keyword">if</span> <span class="keyword">not</span> a%b:</span><br><span class="line">      <span class="keyword">for</span> x <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">0</span>, a//b):</span><br><span class="line">        <span class="keyword">if</span> str1[x*b:(x+<span class="number">1</span>)*b] != str2:</span><br><span class="line">          <span class="keyword">return</span> <span class="string">&#x27;&#x27;</span></span><br><span class="line">      <span class="keyword">return</span> str2</span><br><span class="line">    <span class="keyword">else</span>:</span><br><span class="line">      <span class="keyword">for</span> x <span class="keyword">in</span> <span class="built_in">range</span>(b, <span class="number">0</span>, -<span class="number">1</span>):</span><br><span class="line">        <span class="built_in">print</span>(x)</span><br><span class="line">        <span class="keyword">if</span> x==b <span class="keyword">or</span> b%x <span class="keyword">or</span> a%x:</span><br><span class="line">          <span class="keyword">continue</span></span><br><span class="line">        <span class="keyword">for</span> y <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">0</span>, b//x):</span><br><span class="line">          <span class="keyword">if</span> str2[y*x:(y+<span class="number">1</span>)*x] != str2[b-x:b]:</span><br><span class="line">            <span class="keyword">return</span> <span class="string">&#x27;&#x27;</span></span><br><span class="line">        <span class="keyword">for</span> y <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">0</span>, a//x):</span><br><span class="line">          <span class="keyword">if</span> str1[y*x:(y+<span class="number">1</span>)*x] != str2[<span class="number">0</span>:x]:</span><br><span class="line">            <span class="keyword">return</span> <span class="string">&#x27;&#x27;</span></span><br><span class="line">        <span class="keyword">return</span> str2[<span class="number">0</span>:x]</span><br><span class="line"><span class="comment"># 44 ms	13.9 MB</span></span><br></pre></td></tr></table></figure>


<p>官方解答中还给了一种巧妙的解法，如果 str1 + str2 &#x3D;&#x3D; str2 + str1 的话，<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/greatest-common-divisor-of-strings/solution/zi-fu-chuan-de-zui-da-gong-yin-zi-by-leetcode-solu/">可以证明</a>必定存在这样一个字符串，其长度为两个字符串长度的最大公约数。</p>
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