【leetcode】1091. Shortest Path in Binary Matrix

Topics are as follows:

In an N by N square grid, each cell is either empty (0) or blocked (1).

A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that:

• Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
• C_1 is at location (0, 0) (ie. has value grid[0][0])
• C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
• If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).

Return the length of the shortest such clear path from top-left to bottom-right.  If such a path does not exist, return -1.

 

Example 1:

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

Example 2:

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

 

Note:

1. 1 <= grid.length == grid[0].length <= 100
2. grid[r][c] is 0 or 1

Problem-solving ideas: the typical BFS problem. It calculates the minimum to reach each square from the starting point in turn. Here with the visit [i] [j] records the shortest path from the starting point to the (i, j) is in the process of BFS, it may have multiple paths can reach (i, j), so as long as there is a shorter walk method to reach (i, j), it is necessary to (i, j) is added to the queue again.

code show as below:

class Solution(object):
    def shortestPathBinaryMatrix(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        if grid[0][0] == 1:
            return -1
        visit = []
        val = []
        for i in grid:
            visit.append([0] * len(i))
            val.append([0] * len(i))
        visit[0][0] = 1
        queue = [(0,0)]
        direction = [(0,1),(0,-1),(1,0),(1,-1),(-1,1),(-1,-1),(1,1),(1,-1)]
        while len(queue) > 0:
            x,y = queue.pop(0)
            for (i,j) in direction:
                if x + i >= 0 and x+i < len(grid) and y + j >= 0 and y + j < len(grid[0]) \
                    and grid[x + i][y+j] == 0 and (visit[x+i][y+j] == 0 or visit[x+i][y+j] - 1 > visit[x][y]):
                    queue.append((x+i,y+j))
                    visit[x+i][y+j] = visit[x][y] + 1
        return visit[-1][-1] if visit[-1][-1] > 0 else -1

 

Reproduced in: https: //www.cnblogs.com/seyjs/p/11044772.html