16 lines
4.6 KiB
HTML
16 lines
4.6 KiB
HTML
<!DOCTYPE html><html lang="zh-CN"><head><meta charset="utf-8"><meta name="X-UA-Compatible" content="IE=edge"><title> leetcode-the-masseuse-lcci · MarkDown</title><meta name="description" content="leetcode-the-masseuse-lcci - 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"><meta name="generator" content="Hexo 6.0.0"><link rel="alternate" href="/atom.xml" title="MarkDown" type="application/atom+xml">
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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-the-masseuse-lcci</h1><div class="post-meta"><div class="post-time">2020年4月9日</div></div><div class="post-content"><h3 id="面试题-17-16-按摩师"><a href="#面试题-17-16-按摩师" class="headerlink" title="面试题 17.16. 按摩师"></a>面试题 17.16. 按摩师</h3><p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/the-masseuse-lcci/">题目</a></p>
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<p>一开始以为是用递归，想了半天没想出来，偷看了一下答案。答案的思路跟递归类似，假设在当前 $i$ 时刻，$dp[i][0]$ 为当前预约不接的情况下最长预约时间，$dp[i][1]$ 则为接受当前预约的最长预约时间。</p>
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<p>那很显然，由于不能接受相邻两个预约，$dp[i][1] = dp[i-1][0] + nums_i$</p>
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<p>不接受当前预约的话，上一个预约接不接受都可以，$dp[i][0] = max(dp[i-1][0], dp[i-1][1])$</p>
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<p>最后只要比较两种情况即可 $max(dp[i][0], dp[i][1])$</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">massage</span>(<span class="params">self, nums</span>) -> <span class="built_in">int</span>:</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> nums:</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span></span><br><span class="line"> n = <span class="built_in">len</span>(nums)</span><br><span class="line"> not_choose = <span class="number">0</span></span><br><span class="line"> choose = <span class="number">0</span></span><br><span class="line"> <span class="keyword">for</span> n <span class="keyword">in</span> nums:</span><br><span class="line"> not_choose, choose = <span class="built_in">max</span>(not_choose, choose), not_choose+n</span><br><span class="line"> <span class="keyword">return</span> <span class="built_in">max</span>(not_choose, choose)</span><br><span class="line"> <span class="comment"># 52 ms 13.6 MB</span></span><br></pre></td></tr></table></figure>
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<p>这种问题原来有个名字叫<a target="_blank" rel="noopener" href="https://zh.wikipedia.org/wiki/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92">动态规划</a>，上面推导的方程叫<a target="_blank" rel="noopener" href="https://baike.baidu.com/item/%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%E6%96%B9%E7%A8%8B">状态转移方程</a>，可以找找资料来看一下。</p>
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