124 lines
7.8 KiB
HTML
124 lines
7.8 KiB
HTML
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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-876</h1><div class="post-meta"><div class="post-time">2020年3月26日</div></div><div class="post-content"><h3 id="876-__u94FE_u8868_u7684_u4E2D_u95F4_u7ED3_u70B9"><a href="#876-__u94FE_u8868_u7684_u4E2D_u95F4_u7ED3_u70B9" class="headerlink" title="876. 链表的中间结点"></a>876. 链表的中间结点</h3><p><a href="https://leetcode-cn.com/problems/middle-of-the-linked-list/" target="_blank" rel="noopener">题目</a></p>
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<p>思路是遍历一遍得到整个链表，讲每个 node 放进一个 list，就可以通过下标得到中间的。</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></pre></td><td class="code"><pre><span class="line"><span class="comment"># Definition for singly-linked list.</span></span><br><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">middleNode</span><span class="params">(self, head: ListNode)</span> -> ListNode:</span></span><br><span class="line"> l = []</span><br><span class="line"> n = head</span><br><span class="line"> <span class="keyword">while</span> n.next:</span><br><span class="line"> l.append(n)</span><br><span class="line"> n = n.next</span><br><span class="line"> l.append(n)</span><br><span class="line"> <span class="keyword">return</span> l[len(l)//<span class="number">2</span>]</span><br><span class="line"></span><br><span class="line"> <span class="comment">#44 ms 13.7 MB</span></span><br></pre></td></tr></table></figure>
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<p>看官方解答，还有一个骚操作，通过两个速度不一样的指针，一个一次走一步，一个两次走一步，快的走到底时，慢的就在中间了。</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">middleNode</span><span class="params">(self, head: ListNode)</span> -> ListNode:</span></span><br><span class="line"> slow = fast = head</span><br><span class="line"> <span class="keyword">while</span> fast <span class="keyword">and</span> fast.next:</span><br><span class="line"> slow = slow.next</span><br><span class="line"> fast = fast.next.next</span><br><span class="line"> <span class="keyword">return</span> slow</span><br><span class="line"></span><br><span class="line">作者：LeetCode-Solution</span><br><span class="line">链接：https://leetcode-cn.com/problems/middle-of-the-linked-list/solution/lian-biao-de-zhong-jian-jie-dian-by-leetcode-solut/</span><br><span class="line">来源：力扣（LeetCode）</span><br><span class="line">著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。</span><br></pre></td></tr></table></figure>
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