542. 01 Matrix

Problem

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

The distance between two cells sharing a common edge is 1.

https://leetcode.com/problems/01-matrix/

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.

Note: This question is the same as 1765: https://leetcode.com/problems/map-of-highest-peak/

Test Cases

python
class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:

solution_test.py下载完整源码
py
import pytest

from solution import Solution


@pytest.mark.parametrize('mat, expected', [
    ([[0,0,0],[0,1,0],[0,0,0]], [[0,0,0],[0,1,0],[0,0,0]]),
    ([[0,0,0],[0,1,0],[1,1,1]], [[0,0,0],[0,1,0],[1,2,1]]),

    (
        [[0,1,1,0,1,0,1,0,1,0],[1,1,1,0,0,0,1,0,0,1],[0,1,1,1,0,1,1,0,1,1],[1,1,1,1,0,1,1,1,1,0],[1,0,1,1,1,1,1,1,1,1],[1,1,1,1,0,0,1,0,1,1],[1,1,0,1,0,0,1,1,1,1],[1,1,0,1,0,0,1,0,0,0],[0,0,1,0,1,0,1,1,1,0],[1,1,1,1,0,1,1,0,1,1]],
        [[0,1,1,0,1,0,1,0,1,0],[1,2,1,0,0,0,1,0,0,1],[0,1,2,1,0,1,1,0,1,1],[1,1,2,1,0,1,2,1,1,0],[1,0,1,2,1,1,2,1,2,1],[2,1,1,1,0,0,1,0,1,2],[2,1,0,1,0,0,1,1,1,1],[1,1,0,1,0,0,1,0,0,0],[0,0,1,0,1,0,1,1,1,0],[1,1,2,1,0,1,1,0,1,1]]
     ),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, mat, expected):
    assert sol.updateMatrix(mat) == expected

Thoughts

跟 221. Maximal Square 很像，少了对于全一正方形的判定。Problem 221 中的 h(i, j) 和 v(i, j) 分别类似于本题中格子 (i, j) 到上边最近的 0 和左边最近的 0 的距离（如果某个方向没有 0 本题应该为 inf，problem 221 则为连续 1 的个数）。

定义 dp(i, j) 为格子 (i, j) 到任意方向最近的 0 的距离。易知当 mat[i][j] = 0 时：

dp(i,j)=0

当 mat[i][j] = 1 时：

dp(i,j)=1+\min\begin{cases} dp(i-1,j) \\ dp(i+1,j) \\ dp(i,j-1) \\ dp(i,j+1) \end{cases}

边界值 dp(i, -1) = dp(-1, j) = dp(i, n) = dp(m, j) = ∞。

同时看四个方向可能不太方便，可以遍历两次，第一次从上到下、从左到右，看上边和左边；第二次从下到上、从右到左，看下边和右边。

时间复杂度 O(m * n)，附加的空间复杂度 O(1)。

Code

solution.py下载完整源码
py
class Solution:
    def updateMatrix(self, mat: list[list[int]]) -> list[list[int]]:
        min2 = lambda a, b: a if a <= b else b
        m = len(mat)
        n = len(mat[0])
        dp = [[m+n] * n for _ in range(m)]

        for i in range(m):
            for j in range(n):
                if mat[i][j]:
                    if i > 0: dp[i][j] = min2(dp[i][j], 1 + dp[i-1][j])
                    if j > 0: dp[i][j] = min2(dp[i][j], 1 + dp[i][j-1])
                else:
                    dp[i][j] = 0

        for i in range(m - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                if mat[i][j]:
                    if i < m - 1: dp[i][j] = min2(dp[i][j], 1 + dp[i+1][j])
                    if j < n - 1: dp[i][j] = min2(dp[i][j], 1 + dp[i][j+1])

        return dp

#Problem

#Test Cases

#Thoughts

#Code

Problem

Test Cases

Thoughts

Code