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38 lines
1.4 KiB
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38 lines
1.4 KiB
HTML
<p>Given an <code>m x n</code> integer matrix <code>matrix</code>, if an element is <code>0</code>, set its entire row and column to <code>0</code>'s.</p>
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<p>You must do it <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank">in place</a>.</p>
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<p> </p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg" style="width: 450px; height: 169px;" />
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<pre>
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<strong>Input:</strong> matrix = [[1,1,1],[1,0,1],[1,1,1]]
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<strong>Output:</strong> [[1,0,1],[0,0,0],[1,0,1]]
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</pre>
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<p><strong>Example 2:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg" style="width: 450px; height: 137px;" />
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<pre>
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<strong>Input:</strong> matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
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<strong>Output:</strong> [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>m == matrix.length</code></li>
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<li><code>n == matrix[0].length</code></li>
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<li><code>1 <= m, n <= 200</code></li>
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<li><code>-2<sup>31</sup> <= matrix[i][j] <= 2<sup>31</sup> - 1</code></li>
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</ul>
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<p> </p>
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<p><strong>Follow up:</strong></p>
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<ul>
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<li>A straightforward solution using <code>O(mn)</code> space is probably a bad idea.</li>
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<li>A simple improvement uses <code>O(m + n)</code> space, but still not the best solution.</li>
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<li>Could you devise a constant space solution?</li>
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</ul>
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