01-matrix

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Algorithms

Given an <code>m x n</code> binary matrix <code>mat</code>, return the distance of the nearest <code>0</code> for each cell.

The distance between two cells sharing a common edge is <code>1</code>.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-1-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
Input: mat = [[0,0,0],[0,1,0],[0,0,0]]
Output: [[0,0,0],[0,1,0],[0,0,0]]
</pre>

Example 2:
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-2-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
Input: mat = [[0,0,0],[0,1,0],[1,1,1]]
Output: [[0,0,0],[0,1,0],[1,2,1]]
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 104</code></li>
	<li><code>1 &lt;= m * n &lt;= 104</code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one <code>0</code> in <code>mat</code>.</li>
</ul>

&nbsp;
Note: This question is the same as 1765: <a href="https://leetcode.com/problems/map-of-highest-peak/description/" target="_blank">https://leetcode.com/problems/map-of-highest-peak/</a>

updateMatrix

01 Matrix

Shortest Path to Get Food

shortest-path-to-get-food

获取食物的最短路径

Minimum Operations to Remove Adjacent Ones in Matrix

minimum-operations-to-remove-adjacent-ones-in-matrix

使矩阵中的 1 互不相邻的最小操作数

Difference Between Ones and Zeros in Row and Column

difference-between-ones-and-zeros-in-row-and-column

行和列中一和零的差值

Breadth-First Search

Array

Dynamic Programming

Matrix

给定一个由 <code>0</code> 和 <code>1</code> 组成的矩阵 <code>mat</code> ，请输出一个大小相同的矩阵，其中每一个格子是 <code>mat</code> 中对应位置元素到最近的 <code>0</code> 的距离。

两个相邻元素间的距离为 <code>1</code> 。

 

示例 1：

<img alt="" src="https://pic.leetcode-cn.com/1626667201-NCWmuP-image.png" style="width: 150px; " />

<pre>
输入：mat = [[0,0,0],[0,1,0],[0,0,0]]
输出：[[0,0,0],[0,1,0],[0,0,0]]
</pre>

示例 2：

<img alt="" src="https://pic.leetcode-cn.com/1626667205-xFxIeK-image.png" style="width: 150px; " />

<pre>
输入：mat = [[0,0,0],[0,1,0],[1,1,1]]
输出：[[0,0,0],[0,1,0],[1,2,1]]
</pre>

 

提示：

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 <= m, n <= 104</code></li>
	<li><code>1 <= m * n <= 104</code></li>
	<li><code>mat[i][j] is either 0 or 1.</code></li>
	<li><code>mat</code> 中至少有一个 <code>0 </code></li>
</ul>