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128 lines
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<!DOCTYPE html><html lang="zh-CN"><head><meta charset="utf-8"><meta name="X-UA-Compatible" content="IE=edge"><title> leetcode-206 · MarkDown</title><meta name="description" content="leetcode-206 - Ching"><meta name="viewport" content="width=device-width, initial-scale=1"><link rel="short icon" href="/favicon.png"><link rel="stylesheet" href="/css/apollo.css"><link rel="stylesheet" href="https://fonts.googleapis.com/css?family=Source+Sans+Pro:400,600" type="text/css"><style><!-- hexo-inject:begin --><!-- hexo-inject:end -->mjx-container[jax="SVG"] {
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</style><!-- hexo-inject:begin --><!-- hexo-inject:end --></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="https://bearmiebear.blogspot.com" target="_blank" class="nav-list-link">BEAR</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-206</h1><div class="post-meta"><div class="post-time">2020年3月18日</div></div><div class="post-content"><h3 id="206-__u53CD_u8F6C_u94FE_u8868"><a href="#206-__u53CD_u8F6C_u94FE_u8868" class="headerlink" title="206. 反转链表"></a>206. 反转链表</h3><p><a href="https://leetcode-cn.com/problems/reverse-linked-list/" target="_blank" rel="noopener">题目</a></p>
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<p>最简单的思路是遍历链表一个列表去做存储，通过倒序读取列表的同时改写链表。</p>
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<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></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">ListNode</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">__init__</span><span class="params">(self, x)</span>:</span></span><br><span class="line"> self.val = x</span><br><span class="line"> self.next = <span class="keyword">None</span></span><br><span class="line"></span><br><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">reverseList</span><span class="params">(self, head)</span>:</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> head <span class="keyword">or</span> <span class="keyword">not</span> head.next:</span><br><span class="line"> <span class="keyword">return</span> head</span><br><span class="line"> nl = []</span><br><span class="line"> <span class="keyword">while</span> head.next:</span><br><span class="line"> nl.append(head)</span><br><span class="line"> head = head.next</span><br><span class="line"> nl.append(head)</span><br><span class="line"> l = len(nl)</span><br><span class="line"> <span class="keyword">for</span> x <span class="keyword">in</span> range(l):</span><br><span class="line"> <span class="keyword">if</span> x == <span class="number">0</span>:</span><br><span class="line"> nl[x].next = <span class="keyword">None</span></span><br><span class="line"> <span class="keyword">continue</span></span><br><span class="line"> nl[x].next = nl[x<span class="number">-1</span>]</span><br><span class="line"> <span class="keyword">if</span> x == (l - <span class="number">1</span>):</span><br><span class="line"> <span class="keyword">return</span> nl[x]</span><br></pre></td></tr></table></figure>
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<p>仔细想想自己又傻逼了，何必要遍历两次呢，在第一遍遍历的同时就能操作了：</p>
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<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></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">reverseList</span><span class="params">(self, head)</span>:</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> head <span class="keyword">or</span> <span class="keyword">not</span> head.next:</span><br><span class="line"> <span class="keyword">return</span> head</span><br><span class="line"> prev_node = <span class="keyword">None</span></span><br><span class="line"> next_node = head.next</span><br><span class="line"> <span class="keyword">while</span> head:</span><br><span class="line"> next_node = head.next</span><br><span class="line"> head.next = prev_node</span><br><span class="line"> prev_node = head</span><br><span class="line"> head = next_node</span><br><span class="line"> <span class="keyword">return</span> prev_node</span><br></pre></td></tr></table></figure>
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<p>然后是递归的做法，主要思路是一直进到最深一层–也就是链表的最后一个–的时候开始返回，同时修改那一层的两个 node。一开始踩了一个坑是返回了每一个node，结果最后回到第一层的时候得到的是链表的末端，其实只需要修改链表，并不需要返回 node，所以一开始到达链表末端的时候直接返回那一个node就可以了。</p>
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<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></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">reverseList</span><span class="params">(self, head)</span>:</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> head:</span><br><span class="line"> <span class="keyword">return</span> head</span><br><span class="line"> <span class="keyword">if</span> head.next:</span><br><span class="line"> ss = Solution()</span><br><span class="line"> last = ss.reverseList(head.next)</span><br><span class="line"> head.next.next = head</span><br><span class="line"> head.next = <span class="keyword">None</span></span><br><span class="line"> <span class="keyword">return</span> last</span><br><span class="line"> <span class="keyword">return</span> head</span><br></pre></td></tr></table></figure>
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<p>一开始是用list来打草稿，不过想明白递归之后就大同小异了：</p>
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<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></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">a</span><span class="params">(l:list)</span>->list:</span></span><br><span class="line"> k=[l[<span class="number">0</span>]]</span><br><span class="line"></span><br><span class="line"> <span class="keyword">if</span> l[<span class="number">1</span>:]:</span><br><span class="line"> b=a(l[<span class="number">1</span>:])</span><br><span class="line"> b.extend(k)</span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> <span class="keyword">return</span> [l[<span class="number">0</span>]]</span><br><span class="line"> <span class="keyword">return</span> b</span><br></pre></td></tr></table></figure>
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</div></article></div></section><footer><div class="paginator"><a href="/2020/03/23/leetcode-169/" class="prev">PRVE</a><a href="/2020/03/17/leetcode-121/" class="next">NEXT</a></div><div class="copyright"><p>© 2016 - 2020 <a href="http://blog.tunpok.com">Ching</a>, unless otherwise noted.</p></div></footer><script src="https://cdn.bootcss.com/mathjax/2.5.3/MathJax.js?config=TeX-AMS-MML_HTMLorMML"></script><!-- hexo-inject:begin --><!-- hexo-inject:end --></body></html> |