Leetcode 1905. Count Sub Islands

Tyan

发布于 2021-08-13 11:56:07

4070

发布于 2021-08-13 11:56:07

文章被收录于专栏：SnailTyanSnailTyan

1. Description

2. Solution

**解析：**Version 1，以第二个矩阵中碰到的1作为起点，然后使用广度优先搜索找到所有相邻的1，即一个岛屿，并将所有岛的坐标保存到队列中(值为1的坐标)，将矩阵二中搜索的点对应的值设为2，防止重复搜索，搜索过程中需要同时检查搜索的点是否是矩阵一种的岛屿，如果不是，将标志位设为False，最后根据标志位判断是否是矩阵一种的子岛屿。搜索过程其实就是Flood Fill算法。

Version 1

class Solution:
    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        m = len(grid1)
        n = len(grid1[0])
        count = 0
        queue = collections.deque()
        for i in range(m):
            for j in range(n):
                if grid2[i][j] == 1:
                    queue.append((i, j))
                    count += self.subIslands(queue, grid2, grid1)     
        return count

    
    def subIslands(self, queue, grid, check):
        m = len(grid)
        n = len(grid[0])
        flag = True
        while queue:
            x, y = queue.popleft()
            if check[x][y] == 0:
                flag = False
            if x > 0 and grid[x-1][y] == 1:
                grid[x-1][y] = 2
                queue.append((x-1, y))
            if y > 0 and grid[x][y-1] == 1:
                grid[x][y-1] = 2
                queue.append((x, y-1))
            if x < m-1 and grid[x+1][y] == 1:
                grid[x+1][y] = 2
                queue.append((x+1, y))
            if y < n-1 and grid[x][y+1] == 1:
                grid[x][y+1] = 2
                queue.append((x, y+1))    
        if flag:
            return 1
        else:
            return 0