mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
42 lines
1.6 KiB
HTML
42 lines
1.6 KiB
HTML
|
<p>给定一个 <code><em>m</em> x <em>n</em></code> 的矩阵，如果一个元素为 <strong>0 </strong>，则将其所在行和列的所有元素都设为 <strong>0</strong> 。请使用 <strong><a href="http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95" target="_blank">原地</a></strong> 算法<strong>。</strong></p>
|
|||
|
|
|
|||
|
|
<ul>
|
|||
|
|
</ul>
|
|||
|
|
|
|||
|
|
<p> </p>
|
|||
|
|
|
|||
|
|
<p><strong>示例 1：</strong></p>
|
|||
|
|
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg" style="width: 450px; height: 169px;" />
|
|||
|
|
<pre>
|
|||
|
|
<strong>输入：</strong>matrix = [[1,1,1],[1,0,1],[1,1,1]]
|
|||
|
|
<strong>输出：</strong>[[1,0,1],[0,0,0],[1,0,1]]
|
|||
|
|
</pre>
|
|||
|
|
|
|||
|
|
<p><strong>示例 2：</strong></p>
|
|||
|
|
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg" style="width: 450px; height: 137px;" />
|
|||
|
|
<pre>
|
|||
|
|
<strong>输入：</strong>matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
|
|||
|
|
<strong>输出：</strong>[[0,0,0,0],[0,4,5,0],[0,3,1,0]]
|
|||
|
|
</pre>
|
|||
|
|
|
|||
|
|
<p> </p>
|
|||
|
|
|
|||
|
|
<p><strong>提示：</strong></p>
|
|||
|
|
|
|||
|
|
<ul>
|
|||
|
|
<li><code>m == matrix.length</code></li>
|
|||
|
|
<li><code>n == matrix[0].length</code></li>
|
|||
|
|
<li><code>1 <= m, n <= 200</code></li>
|
|||
|
|
<li><code>-2<sup>31</sup> <= matrix[i][j] <= 2<sup>31</sup> - 1</code></li>
|
|||
|
|
</ul>
|
|||
|
|
|
|||
|
|
<p> </p>
|
|||
|
|
|
|||
|
|
<p><strong>进阶：</strong></p>
|
|||
|
|
|
|||
|
|
<ul>
|
|||
|
|
<li>一个直观的解决方案是使用 <code>O(<em>m</em><em>n</em>)</code> 的额外空间，但这并不是一个好的解决方案。</li>
|
|||
|
|
<li>一个简单的改进方案是使用 <code>O(<em>m</em> + <em>n</em>)</code> 的额外空间，但这仍然不是最好的解决方案。</li>
|
|||
|
|
<li>你能想出一个仅使用常量空间的解决方案吗？</li>
|
|||
|
|
</ul>
|