1905. Count Sub Islands (M)

You are given two m x n binary matrices grid1 and grid2 containing only 0's (representing water) and 1's (representing land). An island is a group of 1's connected 4-directionally (horizontal or vertical). Any cells outside of the grid are considered water cells.

An island in grid2 is considered a sub-island if there is an island in grid1 that contains all the cells that make up this island in grid2.

Return the number of islands in grid2 that are considered sub-islands.

Example 1:

Input: grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]
Output: 3
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are three sub-islands.

Example 2:

Input: grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
Output: 2 
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.

Constraints:

• m == grid1.length == grid2.length

• n == grid1[i].length == grid2[i].length

• 1 <= m, n <= 500

• grid1[i][j] and grid2[i][j] are either 0 or 1.

Solution:

这道题的关键在于，如何快速判断子岛屿？肯定可以借助 Union Find 并查集算法 来判断，不过本文重点在 DFS 算法，就不展开并查集算法了。

什么情况下 grid2 中的一个岛屿 B 是 grid1 中的一个岛屿 A 的子岛？

当岛屿 B 中所有陆地在岛屿 A 中也是陆地的时候，岛屿 B 是岛屿 A 的子岛。

反过来说，如果岛屿 B 中存在一片陆地，在岛屿 A 的对应位置是海水，那么岛屿 B 就不是岛屿 A 的子岛。

那么，我们只要遍历 grid2 中的所有岛屿，把那些不可能是子岛的岛屿排除掉，剩下的就是子岛。

依据这个思路，可以直接写出下面的代码：

int countSubIslands(int[][] grid1, int[][] grid2) {
    int m = grid1.length, n = grid1[0].length;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (grid1[i][j] == 0 && grid2[i][j] == 1) {
                // 这个岛屿肯定不是子岛，淹掉
                dfs(grid2, i, j);
            }
        }
    }
    // 现在 grid2 中剩下的岛屿都是子岛，计算岛屿数量
    int res = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (grid2[i][j] == 1) {
                res++;
                dfs(grid2, i, j);
            }
        }
    }
    return res;
}

// 从 (i, j) 开始，将与之相邻的陆地都变成海水
void dfs(int[][] grid, int i, int j) {
    int m = grid.length, n = grid[0].length;
    if (i < 0 || j < 0 || i >= m || j >= n) {
        return;
    }
    if (grid[i][j] == 0) {
        return;
    }

    grid[i][j] = 0;
    dfs(grid, i + 1, j);
    dfs(grid, i, j + 1);
    dfs(grid, i - 1, j);
    dfs(grid, i, j - 1);
}

这道题的思路和计算「封闭岛屿」数量的思路有些类似，只不过后者排除那些靠边的岛屿，前者排除那些不可能是子岛的岛屿。