1905. 统计子岛屿

1905. 统计子岛屿

java：DFS

class Solution {
    Boolean flag = true;
    private void dfs(int x, int y, int m, int n, int[][] grid1, int[][] grid2) {
        if (x < 0 || x >= m || y < 0 || y >= n || grid2[x][y] != 1) {  // 包含关系：遍历最小范围，如果其中一个点不被大范围的包含，就不是包含关系
            return;
        }
        if (grid1[x][y] != 1) {
            flag = false;
        }
        grid2[x][y] = 0;
        dfs(x - 1, y, m, n, grid1, grid2);
        dfs(x + 1, y, m, n, grid1, grid2);
        dfs(x, y - 1, m, n, grid1, grid2);
        dfs(x, y + 1, m, n, grid1, grid2);
    }

    public int countSubIslands(int[][] grid1, int[][] grid2) {
        int m = grid2.length;
        int n = grid2[0].length;
        int cnt = 0;
        for (int i = 0; i < m; ++i) {
            for(int j = 0; j < n; ++j) {
                if (grid2[i][j] == 1) {
                    flag = true;
                    dfs(i, j, m, n, grid1, grid2);
                    if(flag) {
                        cnt++;
                    }
                }
            }
        }
        return cnt;
    }
}