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39 lines
1.8 KiB
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39 lines
1.8 KiB
HTML
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<p>给你两个大小为 <code>n x n</code> 的二进制矩阵 <code>mat</code> 和 <code>target</code> 。现<strong> 以 90 度顺时针轮转 </strong>矩阵 <code>mat</code> 中的元素 <strong>若干次</strong> ，如果能够使 <code>mat</code> 与 <code>target</code> 一致，返回 <code>true</code> ；否则，返回<em> </em><code>false</code><em> 。</em></p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/05/20/grid3.png" style="width: 301px; height: 121px;" />
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<pre>
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<strong>输入：</strong>mat = [[0,1],[1,0]], target = [[1,0],[0,1]]
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<strong>输出：</strong>true
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<strong>解释：</strong>顺时针轮转 90 度一次可以使 mat 和 target 一致。
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</pre>
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<p><strong>示例 2：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/05/20/grid4.png" style="width: 301px; height: 121px;" />
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<pre>
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<strong>输入：</strong>mat = [[0,1],[1,1]], target = [[1,0],[0,1]]
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<strong>输出：</strong>false
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<strong>解释：</strong>无法通过轮转矩阵中的元素使 equal 与 target 一致。
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</pre>
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<p><strong>示例 3：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/05/26/grid4.png" style="width: 661px; height: 184px;" />
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<pre>
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<strong>输入：</strong>mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]]
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<strong>输出：</strong>true
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<strong>解释：</strong>顺时针轮转 90 度两次可以使 mat 和 target 一致。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>n == mat.length == target.length</code></li>
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<li><code>n == mat[i].length == target[i].length</code></li>
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<li><code>1 <= n <= 10</code></li>
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<li><code>mat[i][j]</code> 和 <code>target[i][j]</code> 不是 <code>0</code> 就是 <code>1</code></li>
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</ul>
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