1091. Shortest Path in Binary Matrix，奇怪的路徑 python 解

 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

這種題目就是給你一個圖，然後要你找最短路徑，而且是完全沒有weight 或是所有weight都一樣

這就是老題目: shorted graph of unweighted graph=bfs (且已經知道起點跟終點啦)

使用bfs 並且找level

最後回傳level在最右下角 即可

參考底下的python code 唷

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        
        #bfs
        n=len(grid)
        if grid[0][0]==1:
            return -1
        visited={(0,0)}
        queue=deque([(0,0)])
        level=dict()
        level[(0,0)]=1
        while queue:
             for i in range(len(queue)):
                node=queue.popleft()
                x,y=node
                for dx,dy in {(-1,0),(1,0),(0,1),(0,-1),(1,1),(1,-1),(-1,1),(-1,-1)}:
                    if x+dx<n and x+dx>=0 and y+dy<n and y+dy>=0 and grid[x+dx][y+dy]==0:
                        if (x+dx,y+dy) in visited:
                            continue
                        visited.add((x+dx,y+dy))
                        queue.append((x+dx,y+dy))
                        level[(x+dx,y+dy)]=level[(x,y)]+1
        
        if (n-1,n-1) in level:
            return level[(n-1,n-1)]
        else:
            return -1