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128 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-01-matrix</h1><div class="post-meta"><div class="post-time">2020年4月16日</div></div><div class="post-content"><h3 id="542-_01__u77E9_u9635"><a href="#542-_01__u77E9_u9635" class="headerlink" title="542. 01 矩阵"></a>542. 01 矩阵</h3><p><a href="https://leetcode-cn.com/problems/01-matrix/" 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>想了两种思路</p>
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<ol>
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<li><p>0 位置的上下左右是 1， 上下左右中有跟 1 相邻的就是 2，以此类推，从 0 的坐标开始往上下左右四个方向扩散。如果我们把同意个距离的看作是一层，可以用一个队列依次存放每一层的坐标，直至每个坐标都被计算过。</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></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">updateMatrix</span><span class="params">(self, matrix: List[List[int]])</span> -> List[List[int]]:</span></span><br><span class="line"> m, n = len(matrix), len(matrix[<span class="number">0</span>])</span><br><span class="line"> dist = [[<span class="number">0</span>] * n <span class="keyword">for</span> _ <span class="keyword">in</span> range(m)]</span><br><span class="line"> zeroes_pos = [(i, j) <span class="keyword">for</span> i <span class="keyword">in</span> range(m) <span class="keyword">for</span> j <span class="keyword">in</span> range(n) <span class="keyword">if</span> matrix[i][j] == <span class="number">0</span>]</span><br><span class="line"> <span class="comment"># 将所有的 0 添加进初始队列中</span></span><br><span class="line"> q = collections.deque(zeroes_pos)</span><br><span class="line"> seen = set(zeroes_pos)</span><br><span class="line"></span><br><span class="line"> <span class="comment"># 广度优先搜索</span></span><br><span class="line"> <span class="keyword">while</span> q:</span><br><span class="line"> i, j = q.popleft()</span><br><span class="line"> <span class="keyword">for</span> ni, nj <span class="keyword">in</span> [(i - <span class="number">1</span>, j), (i + <span class="number">1</span>, j), (i, j - <span class="number">1</span>), (i, j + <span class="number">1</span>)]:</span><br><span class="line"> <span class="keyword">if</span> <span class="number">0</span> <= ni < m <span class="keyword">and</span> <span class="number">0</span> <= nj < n <span class="keyword">and</span> (ni, nj) <span class="keyword">not</span> <span class="keyword">in</span> seen:</span><br><span class="line"> dist[ni][nj] = dist[i][j] + <span class="number">1</span></span><br><span class="line"> q.append((ni, nj))</span><br><span class="line"> seen.add((ni, nj))</span><br><span class="line"></span><br><span class="line"> <span class="keyword">return</span> dist</span><br></pre></td></tr></table></figure>
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</li>
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<li><p>从左上角开往右下角遍历矩阵，当前坐标的距离由左和上两个位置的值确定。遍历一遍后，再反过来从右下角开始往左上角遍历，当前坐标的距离根据右和下两个位置的值确定，比较这两次得出的值中较小的一个即为该点的距离。</p>
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</li>
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</ol>
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</div></article></div></section><footer><div class="paginator"><a href="/2020/04/16/leetcode-merge-intervals/" class="prev">PRVE</a><a href="/2020/04/14/leetcode-add-two-numbers-ii/" 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> |