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122 lines
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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-merge-intervals</h1><div class="post-meta"><div class="post-time">2020年4月16日</div></div><div class="post-content"><h3 id="56-__u5408_u5E76_u533A_u95F4"><a href="#56-__u5408_u5E76_u533A_u95F4" class="headerlink" title="56. 合并区间"></a>56. 合并区间</h3><p><a href="https://leetcode-cn.com/problems/merge-intervals/" 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>首先将区间按起点由小到大排序，这样相邻的两个就能通过终点判断是否重合。</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></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">merge</span><span class="params">(self, intervals)</span>:</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> intervals:</span><br><span class="line"> <span class="keyword">return</span> []</span><br><span class="line"> intervals.sort()</span><br><span class="line"> merged = []</span><br><span class="line"> l = len(intervals)</span><br><span class="line"> m = intervals[<span class="number">0</span>]</span><br><span class="line"> <span class="keyword">for</span> x <span class="keyword">in</span> range(l<span class="number">-1</span>):</span><br><span class="line"> j = intervals[x+<span class="number">1</span>]</span><br><span class="line"> <span class="keyword">if</span> m[<span class="number">1</span>] >= j[<span class="number">0</span>]:</span><br><span class="line"> <span class="keyword">if</span> m[<span class="number">1</span>] <= j[<span class="number">1</span>]:</span><br><span class="line"> m = [m[<span class="number">0</span>], j[<span class="number">1</span>]]</span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> <span class="keyword">continue</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> merged.append(m)</span><br><span class="line"> m = j</span><br><span class="line"> <span class="keyword">if</span> m:</span><br><span class="line"> merged.append(m)</span><br><span class="line"> <span class="keyword">return</span> merged</span><br></pre></td></tr></table></figure>
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