1091. Shortest Path in Binary Matrix

BFS

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

分析

要求找左上到右下最短距离，八个方向，只能使用0

错误：

step不要改，直接传入q时step+1，否则会影响该子层别的节点

需要visited !!!!!

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        q = deque()
        n, m = len(grid), len(grid[0])
        if grid[0][0] == 1:
            return -1
        q.append((0, 0, 1))
        grid[0][0] = 1
        dir = [(-1, 0), (1, 0), (0, -1), (0, 1),(-1,-1),(-1,1),(1,-1),(1,1)]
        while q:
            x,y,step = q.popleft()
            if x == n - 1 and y == m - 1:
                return step
            for dx, dy in dir:
                nx, ny = x + dx, y + dy
                if 0<=nx<n and 0<=ny<m and grid[nx][ny] == 0:
                    q.append((nx, ny,step+1))
                    grid[nx][ny] = 1
                
        return -1