Leetcode 1905. Count Sub Islands

文章作者：Tyan
博客：noahsnail.com | CSDN | 简书

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

Reference

https://leetcode.com/problems/count-sub-islands/

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Leetcode 1905. Count Sub Islands

1. Description

2. Solution

Reference

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