LeetCode 542. 01 Matrix 中离最近0的距离

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本文探讨了在由0和1组成的矩阵中寻找每个单元格到最近0元素的距离的算法。通过深度优先搜索（DFS）和广度优先搜索（BFS）的方法，详细解释了如何有效地解决这个问题，并提供了一个Python实现的示例。此外，还提到了动态规划（DP）作为一种替代解决方案。

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1:

Input:
[[0,0,0],
 [0,1,0],
 [0,0,0]]

Output:
[[0,0,0],
 [0,1,0],
 [0,0,0]]

Example 2:

Input:
[[0,0,0],
 [0,1,0],
 [1,1,1]]

Output:
[[0,0,0],
 [0,1,0],
 [1,2,1]]

Note:

The number of elements of the given matrix will not exceed 10,000.
There are at least one 0 in the given matrix.
The cells are adjacent in only four directions: up, down, left and right.

------------------------------------------------------------------------

Nice problem!!! DFS first but TLE due to dup check. Then BFS pass the problem:

import sys

class Solution:
    def updateMatrix(self, matrix: List[List[int]]) -> List[List[int]]:
        rows,cols = len(matrix),len(matrix[0])
        res = [[0 if matrix[i][j] == 0 else sys.maxsize for j in range(cols)] for i in range(rows)]
        vis = [(i,j) for j in range(cols) for i in range(rows) if matrix[i][j] == 0 ]
        layers = [set(vis),set()]
        c,n,dis = 0,1,0
        while (layers[c]):
            for (x,y) in layers[c]:
                for (dx,dy) in [(0,1),(1,0),(0,-1),(-1,0)]:
                    nx,ny = x+dx,y+dy
                    if (nx>=0 and nx<rows and ny>=0 and ny<cols and dis+1<res[nx][ny]):
                        res[nx][ny] = dis+1
                        layers[n].add((nx,ny))
            layers[c].clear()
            n,c = c,n
            dis += 1
        return res

DP is also a nice solution! From left up to right down in the first round, then from right down to left up for the 2nd round!!!