determine-whether-matrix-can-be-obtained-by-rotation

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Algorithms

Given two <code>n x n</code> binary matrices <code>mat</code> and <code>target</code>, return <code>true</code> if it is possible to make <code>mat</code> equal to <code>target</code> by rotating <code>mat</code> in 90-degree increments, or <code>false</code> otherwise.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/20/grid3.png" style="width: 301px; height: 121px;" />
<pre>
Input: mat = [[0,1],[1,0]], target = [[1,0],[0,1]]
Output: true
Explanation: We can rotate mat 90 degrees clockwise to make mat equal target.
</pre>

Example 2:
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/20/grid4.png" style="width: 301px; height: 121px;" />
<pre>
Input: mat = [[0,1],[1,1]], target = [[1,0],[0,1]]
Output: false
Explanation: It is impossible to make mat equal to target by rotating mat.
</pre>

Example 3:
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/26/grid4.png" style="width: 661px; height: 184px;" />
<pre>
Input: mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]]
Output: true
Explanation: We can rotate mat 90 degrees clockwise two times to make mat equal target.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>n == mat.length == target.length</code></li>
	<li><code>n == mat[i].length == target[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 10</code></li>
	<li><code>mat[i][j]</code> and <code>target[i][j]</code> are either <code>0</code> or <code>1</code>.</li>
</ul>

findRotation

Determine Whether Matrix Can Be Obtained By Rotation

Rotate Image

rotate-image

旋转图像

Array

Matrix

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 以 90 度顺时针轮转 矩阵 <code>mat</code> 中的元素 若干次 ，如果能够使 <code>mat</code> 与 <code>target</code> 一致，返回 <code>true</code> ；否则，返回 <code>false</code> 。

 

示例 1：
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/20/grid3.png" style="width: 301px; height: 121px;" />
<pre>
输入：mat = [[0,1],[1,0]], target = [[1,0],[0,1]]
输出：true
解释：顺时针轮转 90 度一次可以使 mat 和 target 一致。
</pre>

示例 2：
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/20/grid4.png" style="width: 301px; height: 121px;" />
<pre>
输入：mat = [[0,1],[1,1]], target = [[1,0],[0,1]]
输出：false
解释：无法通过轮转矩阵中的元素使 equal 与 target 一致。
</pre>

示例 3：
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/26/grid4.png" style="width: 661px; height: 184px;" />
<pre>
输入：mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]]
输出：true
解释：顺时针轮转 90 度两次可以使 mat 和 target 一致。
</pre>

 

提示：

<ul>
	<li><code>n == mat.length == target.length</code></li>
	<li><code>n == mat[i].length == target[i].length</code></li>
	<li><code>1 <= n <= 10</code></li>
	<li><code>mat[i][j]</code> 和 <code>target[i][j]</code> 不是 <code>0</code> 就是 <code>1</code></li>
</ul>