mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
43 lines
1.6 KiB
HTML
43 lines
1.6 KiB
HTML
<p>Given a <code>m x n</code> binary matrix <code>mat</code>. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing <code>1</code> to <code>0</code> and <code>0</code> to <code>1</code>). A pair of cells are called neighbors if they share one edge.</p>
|
|
|
|
<p>Return the <em>minimum number of steps</em> required to convert <code>mat</code> to a zero matrix or <code>-1</code> if you cannot.</p>
|
|
|
|
<p>A <strong>binary matrix</strong> is a matrix with all cells equal to <code>0</code> or <code>1</code> only.</p>
|
|
|
|
<p>A <strong>zero matrix</strong> is a matrix with all cells equal to <code>0</code>.</p>
|
|
|
|
<p> </p>
|
|
<p><strong>Example 1:</strong></p>
|
|
<img alt="" src="https://assets.leetcode.com/uploads/2019/11/28/matrix.png" style="width: 409px; height: 86px;" />
|
|
<pre>
|
|
<strong>Input:</strong> mat = [[0,0],[0,1]]
|
|
<strong>Output:</strong> 3
|
|
<strong>Explanation:</strong> One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.
|
|
</pre>
|
|
|
|
<p><strong>Example 2:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> mat = [[0]]
|
|
<strong>Output:</strong> 0
|
|
<strong>Explanation:</strong> Given matrix is a zero matrix. We do not need to change it.
|
|
</pre>
|
|
|
|
<p><strong>Example 3:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> mat = [[1,0,0],[1,0,0]]
|
|
<strong>Output:</strong> -1
|
|
<strong>Explanation:</strong> Given matrix cannot be a zero matrix.
|
|
</pre>
|
|
|
|
<p> </p>
|
|
<p><strong>Constraints:</strong></p>
|
|
|
|
<ul>
|
|
<li><code>m == mat.length</code></li>
|
|
<li><code>n == mat[i].length</code></li>
|
|
<li><code>1 <= m, n <= 3</code></li>
|
|
<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
|
|
</ul>
|