[leetcode] 542. 01 Matrix

题目

• Given an m x n binary matrix mat, return the distance of the nearest 0 for each cell.

  The distance between two adjacent cells is 1.

  Example 1:

Input: mat = [[0,0,0],[0,1,0],[0,0,0]]
Output: [[0,0,0],[0,1,0],[0,0,0]]

Example 2:

Input: mat = [[0,0,0],[0,1,0],[1,1,1]]
Output: [[0,0,0],[0,1,0],[1,2,1]]

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 104
• 1 <= m * n <= 104
• mat[i][j] is either 0 or 1.
• There is at least one 0 in mat.

思路

bfs，以0号格为中心向四周扩展并更新距离值，由于是广度优先，遍历到的距离必定是最小的，不必为已遍历的格比大小。
动态规划，从左上角向右下角进行动态规划，再从右小角向左上角动态规划。

代码

python版本：

import queue
class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
        distances = [[9999999 for _ in range(len(mat[0]))] for _ in range(len(mat))]
        q = queue.Queue()
        [q.put((i, j, 0,)) for i in range(len(mat)) for j in range(len(mat[i])) if mat[i][j] == 0]
        while not q.empty():
            i, j, dis = q.get()
            if i<0 or i>=len(distances) or j<0 or j>=len(distances[0]) or distances[i][j]<=dis:
                continue
            distances[i][j] = dis
            q.put((i, j+1, distances[i][j]+1,))
            q.put((i, j-1, distances[i][j]+1,))
            q.put((i+1, j, distances[i][j]+1,))
            q.put((i-1, j, distances[i][j]+1,))
        return distances

# 动态规划
class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
        for i in range(len(mat)):
            for j in range(len(mat[i])):
                if mat[i][j] > 0:
                    up = mat[i-1][j] if i > 0 else math.inf
                    left = mat[i][j-1] if j > 0 else math.inf
                    mat[i][j] = min(up, left)+1
        for i in range(len(mat)-1, -1, -1):
            for j in range(len(mat[i])-1, -1, -1):
                if mat[i][j] > 0:
                    down = mat[i+1][j] if i+1 < len(mat) else math.inf
                    right = mat[i][j+1] if j+1 < len(mat[i]) else math.inf
                    mat[i][j] = min(mat[i][j], down+1, right+1)
        return mat