125 lines
8.8 KiB
HTML
125 lines
8.8 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-121</h1><div class="post-meta"><div class="post-time">2020年3月17日</div></div><div class="post-content"><h3 id="121-__u4E70_u5356_u80A1_u7968_u7684_u6700_u4F73_u65F6_u673A"><a href="#121-__u4E70_u5356_u80A1_u7968_u7684_u6700_u4F73_u65F6_u673A" class="headerlink" title="121. 买卖股票的最佳时机"></a>121. 买卖股票的最佳时机</h3><p><a href="https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/" target="_blank" rel="noopener">题目</a></p>
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<!-- hexo-inject:begin --><!-- hexo-inject:end --><a id="more"></a>
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<p>原始答案：</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">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">maxProfit</span><span class="params">(self, prices)</span> -> int:</span></span><br><span class="line"> profit = <span class="number">0</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> prices:</span><br><span class="line"> <span class="keyword">return</span> profit</span><br><span class="line"> min_buyin = prices[<span class="number">0</span>]</span><br><span class="line"> max_sellout = prices[<span class="number">0</span>]</span><br><span class="line"> l = len(prices)</span><br><span class="line"> i = <span class="number">0</span></span><br><span class="line"> <span class="keyword">for</span> buyin <span class="keyword">in</span> prices:</span><br><span class="line"> <span class="keyword">if</span> i == l:</span><br><span class="line"> <span class="keyword">return</span> profit</span><br><span class="line"> <span class="keyword">if</span> min_buyin <= buyin:</span><br><span class="line"> max_sellout = max(prices[i:])</span><br><span class="line"> p = max_sellout - min_buyin</span><br><span class="line"> <span class="keyword">if</span> p > profit:</span><br><span class="line"> profit = p</span><br><span class="line"> <span class="keyword">if</span> buyin < min_buyin:</span><br><span class="line"> min_buyin = buyin</span><br><span class="line"> i += <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> profit</span><br><span class="line"><span class="comment">#1880 ms 14.4 MB</span></span><br></pre></td></tr></table></figure>
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<p>主要思路是找到波谷，如果当前价格比前一天要低，则还是在去往波谷的路上；当价格比前一天高或相同时，则到达了一个波谷，计算波谷和之后的波峰的差，就是这一段的利润。将从头至尾过一次，就能找到所有波谷和其后波峰的差，返回最大的即可。但是这个明显地在重复max，时间复杂度是O(n^2)，看起来就很傻逼。</p>
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<p>仔细想想，其实并不需要直接找出波谷后的波峰，只要在for loop时保持波谷为最低的那个，就能算出每一个后续与波谷的差，找最大差即可。改了下代码变成这样</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">maxProfit</span><span class="params">(self, prices)</span> -> int:</span></span><br><span class="line"> profit = <span class="number">0</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> prices:</span><br><span class="line"> <span class="keyword">return</span> profit</span><br><span class="line"> min_buyin = prices[<span class="number">0</span>]</span><br><span class="line"> <span class="keyword">for</span> i <span class="keyword">in</span> range(<span class="number">1</span>, len(prices)):</span><br><span class="line"> min_buyin = min(min_buyin, prices[i<span class="number">-1</span>])</span><br><span class="line"> profit = max(prices[i] - min_buyin, profit)</span><br><span class="line"> <span class="keyword">return</span> profit</span><br><span class="line"> <span class="comment">#44 ms 14.4 MB</span></span><br></pre></td></tr></table></figure>
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