動機

找最短路徑，bfs 但是沒想到最後死在判別重複上

Problem

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

Sol

用set直接超時，如果把看過得改成1就直接過了

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        if grid[0][0] == 1 or grid[-1][-1] == 1:
            return -1
        else:
            q = deque([[(0,0), 1]])
            goal = (len(grid)-1, len(grid[0])-1)
            while q:
                (i,j), ret = q.popleft()
                if (i,j) == goal:
                    return ret
                else:
                    dirs = [(i+1,j),(i,j+1),(i-1,j),(i,j-1),(i+1,j+1),(i+1,j-1),(i-1,j-1),(i-1,j+1)]
                    dirs = [(x,y) for (x,y) in dirs if 0 <= x < len(grid) and 0 <= y < len(grid[i]) and grid[x][y] == 0]
                    for (x,y) in dirs:
                        grid[x][y] = 1
                        q.append([(x,y), ret+1])
            return -1