题目

Given an n x n binary matrix grid, return the length of the shortest clear path in the matrix. If there is no clear path, return -1.

A clear path in a binary matrix is a path from the top-left cell (i.e., (0, 0)) to the bottom-right cell (i.e., (n - 1, n - 1)) such that:

• All the visited cells of the path are 0.
• All the adjacent cells of the path are 8-directionally connected (i.e., they are different and they share an edge or a corner).

The length of a clear path is the number of visited cells of this path.

Example 1:

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

Example 2:

Input: grid = [[0,0,0],[1,1,0],[1,1,0]]
Output: 4

Example 3:

Input: grid = [[1,0,0],[1,1,0],[1,1,0]]
Output: -1

Constraints:

• n == grid.length
• n == grid[i].length
• 1 <= n <= 100
• grid[i][j] is 0 or 1

思路

dfs超时，而bfs却能够顺利通过，看来遇到迷宫题还是优先bfs为好。由于bfs遍历顺利永远是从近到远，不用判断到终点的距离是否最近，第一个到终点的就是距离最短的路径。

代码

python版本：

# dfs，超时
class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        def dfs(i, j):
            if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == 1:
                return 99999999
            if i == len(grid)-1 and j == len(grid[0])-1:
                return 1
            res = 99999999
            for x, y in [[1, 0], [-1, 0], [0, 1], [0, -1], [1, 1], [1, -1], [-1, 1], [-1, -1]]:
                grid[i][j] = 1
                res = min(res, dfs(i+x, j+y))
                grid[i][j] = 0

            return 1+res
        res = dfs(0, 0)
        return res if res < 99999999 else -1

# bfs
class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        q = [(0, 0, 1)]
        for i, j, d in q:
            if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == 1:
                continue
            if i == len(grid)-1 and j == len(grid[0])-1:
                return d
            grid[i][j] = 1
            for x, y in ((0, 1), (0, -1), (1, 0), (-1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)):
                q.append((i+x, j+y, d+1))
        return -1