BFS 基础写法 —— 以 LeetCode #1091 Shortest Path in Binary Matrix 为例

Posted 2021-03-11 sbj123456789

Question

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

解析

简单来说就是经典的BFS路径搜索，从左上角走到右下角，0是通路，1是障碍。

先给出最简洁的写法：

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        n = len(grid)
        if grid[0][0] or grid[n-1][n-1]:
            return -1
        record = [(0,0,1)]
        for i, j, d in record:
            if i == n-1 and j == n-1: return d
            for x, y in ((i-1,j-1),(i-1,j),(i-1,j+1),(i,j-1),(i,j+1),(i+1,j-1),(i+1,j),(i+1,j+1)):
                if 0 <= x < n and 0 <= y < n and not grid[x][y]:
                    record.append((x, y, d+1))
                    grid[x][y] = 1
        return -1

用 record 这个list来记录状态，然后直接一边在后面不停加上下一层的记录，一边iterate

注意到 由于是BFS，采用将已经过的grid置为1来避免重复访问（BFS才能这样做）

即便如此简洁，此方法还是有一些问题，因此更好的方法为：

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        if not grid or grid[0][0] or grid[-1][-1]: 
            return -1
        
        m, n = len(grid), len(grid[0])
        front, end, level = set([(0, 0)]), set([(m-1,n-1)]), 1
        
        while front:

            # intersection
            if front & end: 
                return level
            
            for x, y in front: 
                grid[x][y] = 1
            
            front = set((x+dx, y+dy) for x, y in front for dx, dy in [(1,1),(1,-1),(-1,1),(-1,-1),(0,1),(0,-1),(-1,0),(1,0)] if 0<=x+dx<m and 0<=y+dy<n and grid[x+dx][y+dy]==0)
            
            level += 1
            
            if len(front) > len(end): 
                front, end = end, front
        return -1
        

此方法解决了两个问题：

1. 更新每一层存的状态，减小内存空间

2. level只需一个变量存就够了（因为是BFS）

注意其采用了set而不用list，可以减少重复情况带来的重复计算

 

参考：

https://leetcode.com/problems/shortest-path-in-binary-matrix/discuss/312827/Python-Concise-BFS