0542. 01 矩阵【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - 多源 BFS

1. 📝 题目描述

• leetcode

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

两个相邻元素间的距离为 1。

---

示例 1：

输入：mat = [[0,0,0],[0,1,0],[0,0,0]]
输出：[[0,0,0],[0,1,0],[0,0,0]]

示例 2：

输入：mat = [[0,0,0],[0,1,0],[1,1,1]]
输出：[[0,0,0],[0,1,0],[1,2,1]]

---

提示：

• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 10^4
• 1 <= m * n <= 10^4
• mat[i][j] is either 0 or 1.
• mat 中至少有一个 0

2. 🎯 s.1 - 多源 BFS

c

int** updateMatrix(int** mat, int matSize, int* matColSize, int* returnSize, int** returnColumnSizes) {
    int m = matSize, n = matColSize[0];
    int** dist = (int**)malloc(sizeof(int*) * m);
    int* queue = (int*)malloc(sizeof(int) * m * n * 2);
    int head = 0, tail = 0;
    for (int i = 0; i < m; i++) {
        dist[i] = (int*)malloc(sizeof(int) * n);
        for (int j = 0; j < n; j++) {
            if (mat[i][j] == 0) {
                dist[i][j] = 0;
                queue[tail++] = i;
                queue[tail++] = j;
            } else {
                dist[i][j] = m + n;
            }
        }
    }
    int dirs[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
    while (head < tail) {
        int x = queue[head++], y = queue[head++];
        for (int d = 0; d < 4; d++) {
            int nx = x + dirs[d][0], ny = y + dirs[d][1];
            if (nx >= 0 && nx < m && ny >= 0 && ny < n && dist[nx][ny] > dist[x][y] + 1) {
                dist[nx][ny] = dist[x][y] + 1;
                queue[tail++] = nx;
                queue[tail++] = ny;
            }
        }
    }
    free(queue);
    *returnSize = m;
    *returnColumnSizes = (int*)malloc(sizeof(int) * m);
    for (int i = 0; i < m; i++) (*returnColumnSizes)[i] = n;
    return dist;
}

js

/**
 * @param {number[][]} mat
 * @return {number[][]}
 */
var updateMatrix = function (mat) {
  const m = mat.length,
    n = mat[0].length
  const dist = Array.from({ length: m }, () => new Array(n).fill(Infinity))
  const queue = []
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (mat[i][j] === 0) {
        dist[i][j] = 0
        queue.push([i, j])
      }
    }
  }
  const dirs = [
    [1, 0],
    [-1, 0],
    [0, 1],
    [0, -1],
  ]
  let head = 0
  while (head < queue.length) {
    const [x, y] = queue[head++]
    for (const [dx, dy] of dirs) {
      const nx = x + dx,
        ny = y + dy
      if (
        nx >= 0 &&
        nx < m &&
        ny >= 0 &&
        ny < n &&
        dist[nx][ny] > dist[x][y] + 1
      ) {
        dist[nx][ny] = dist[x][y] + 1
        queue.push([nx, ny])
      }
    }
  }
  return dist
}

py

class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
        m, n = len(mat), len(mat[0])
        dist = [[float('inf')] * n for _ in range(m)]
        queue = deque()
        for i in range(m):
            for j in range(n):
                if mat[i][j] == 0:
                    dist[i][j] = 0
                    queue.append((i, j))
        while queue:
            x, y = queue.popleft()
            for dx, dy in ((1,0),(-1,0),(0,1),(0,-1)):
                nx, ny = x + dx, y + dy
                if 0 <= nx < m and 0 <= ny < n and dist[nx][ny] > dist[x][y] + 1:
                    dist[nx][ny] = dist[x][y] + 1
                    queue.append((nx, ny))
        return dist

• 时间复杂度：$O(mn)$，其中 m 和 n 是矩阵的行数和列数
• 空间复杂度：$O(mn)$

算法思路：

• 将所有 0 元素作为起点加入队列，距离初始化为 0
• BFS 逐层向外扩展，更新每个 1 元素到最近 0 的距离