README.md

1905. 统计子岛屿

English Version

题目描述

给你两个 m x n 的二进制矩阵 grid1 和 grid2 ，它们只包含 0 （表示水域）和 1 （表示陆地）。一个 岛屿 是由 四个方向 （水平或者竖直）上相邻的 1 组成的区域。任何矩阵以外的区域都视为水域。

如果 grid2 的一个岛屿，被 grid1 的一个岛屿 完全 包含，也就是说 grid2 中该岛屿的每一个格子都被 grid1 中同一个岛屿完全包含，那么我们称 grid2 中的这个岛屿为 子岛屿 。

请你返回 grid2 中 子岛屿 的 数目 。

 

示例 1：

输入：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]]
输出：3
解释：如上图所示，左边为 grid1 ，右边为 grid2 。
grid2 中标红的 1 区域是子岛屿，总共有 3 个子岛屿。

示例 2：

输入：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]]
输出：2 
解释：如上图所示，左边为 grid1 ，右边为 grid2 。
grid2 中标红的 1 区域是子岛屿，总共有 2 个子岛屿。

 

提示：

• m == grid1.length == grid2.length
• n == grid1[i].length == grid2[i].length
• 1 <= m, n <= 500
• grid1[i][j] 和 grid2[i][j] 都要么是 0 要么是 1 。

解法

深度优先搜索。

Python3

class Solution:
    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        def dfs(grid1, grid2, i, j, m, n) -> bool:
            res = grid1[i][j] == 1
            grid2[i][j] = 0
            for x, y in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
                a, b = i + x, j + y
                if a >= 0 and a < m and b >= 0 and b < n and grid2[a][b] == 1:
                    if not dfs(grid1, grid2, a, b, m, n):
                        res = False
            return res

        m, n = len(grid1), len(grid1[0])
        count = 0
        for i in range(m):
            for j in range(n):
                if grid2[i][j] == 1 and dfs(grid1, grid2, i, j, m, n):
                    count += 1
        return count

Java

class Solution {
    private int[][] directions = {{0, 1}, {0, - 1}, {1, 0}, {-1, 0}};

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

    private boolean dfs(int[][] grid1, int[][] grid2, int i, int j, int m, int n) {
        boolean res = grid1[i][j] == 1;
        grid2[i][j] = 0;

        for (int[] direction : directions) {
            int a = i + direction[0], b = j + direction[1];
            if (a >= 0 && a < m && b >= 0 && b < n && grid2[a][b] == 1) {
                if (!dfs(grid1, grid2, a, b, m, n)) {
                    res = false;
                }
            }
        }
        return res;
    }
}

TypeScript

function countSubIslands(grid1: number[][], grid2: number[][]): number {
    let m = grid1.length, n = grid1[0].length;
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid2[i][j] == 1 && dfs(grid1, grid2, i, j)) {
                ++ans;
            }
        }
    }
    return ans;
};

function dfs(grid1: number[][], grid2: number[][], i: number, j: number): boolean {
    let m = grid1.length, n = grid1[0].length;
    let ans = true;
    if (grid1[i][j] == 0) {
        ans = false;
    }
    grid2[i][j] = 0;
    for (let [dx, dy] of [[0, 1], [0, -1], [1, 0], [-1, 0]]) {
        let x = i + dx, y = j + dy;
        if (x < 0 || x > m - 1 || y < 0 || y > n - 1 || grid2[x][y] == 0) {
            continue;
        }
        if (!dfs(grid1, grid2, x, y)) {
            ans = false;
        }
    }
    return ans;
}

C++

class Solution {
public:
    int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
        int m = grid1.size(), n = grid1[0].size();
        int count = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid2[i][j] == 1 && dfs(grid1, grid2, i, j, m, n)) {
                    ++count;
                }
            }
        }
        return count;
    }

private:
    vector<vector<int>> directions = {{0, 1}, {0, - 1}, {1, 0}, {-1, 0}};
    bool dfs(vector<vector<int>>& grid1, vector<vector<int>>& grid2, int i, int j, int m, int n) {
        bool res = grid1[i][j] == 1;
        grid2[i][j] = 0;

        for (auto direction : directions) {
            int a = i + direction[0], b = j + direction[1];
            if (a >= 0 && a < m && b >= 0 && b < n && grid2[a][b] == 1) {
                if (!dfs(grid1, grid2, a, b, m, n)) {
                    res = false;
                }
            }
        }
        return res;
    }
};

Go

func countSubIslands(grid1 [][]int, grid2 [][]int) int {
	m, n := len(grid1), len(grid1[0])
	count := 0
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid2[i][j] == 1 && dfs(grid1, grid2, i, j, m, n) {
				count++
			}
		}
	}
	return count
}

func dfs(grid1 [][]int, grid2 [][]int, i, j, m, n int) bool {
	res := grid1[i][j] == 1
	grid2[i][j] = 0
	directions := [4][2]int{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}
	for _, direction := range directions {
		a, b := i+direction[0], j+direction[1]
		if a >= 0 && a < m && b >= 0 && b < n && grid2[a][b] == 1 {
			if !dfs(grid1, grid2, a, b, m, n) {
				res = false
			}
		}
	}
	return res
}

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