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695. 岛屿的最大面积

题目描述

给你一个大小为 m x n 的二进制矩阵 grid 。

岛屿 是由一些相邻的 1 (代表土地) 构成的组合，这里的「相邻」要求两个 1 必须在 水平或者竖直的四个方向上 相邻。你可以假设 grid 的四个边缘都被 0（代表水）包围着。

岛屿的面积是岛上值为 1 的单元格的数目。

计算并返回 grid 中最大的岛屿面积。如果没有岛屿，则返回面积为 0 。

 

示例 1：

输入：grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
输出：6
解释：答案不应该是 11 ，因为岛屿只能包含水平或垂直这四个方向上的 1 。

示例 2：

输入：grid = [[0,0,0,0,0,0,0,0]]
输出：0

 

提示：

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 50
  • grid[i][j] 为 0 或 1

解法

方法一：DFS

我们可以遍历每一个格子 $(i, j)$，从每个格子开始进行深度优先搜索，如果搜索到的格子是陆地，就将当前格子标记为已访问，并且继续搜索上、下、左、右四个方向的格子。搜索结束后，计算标记的陆地的数量，即为岛屿的面积。我们找出最大的岛屿面积即为答案。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是二维数组的行数和列数。

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class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        def dfs(i: int, j: int) -> int:
            if grid[i][j] == 0:
                return 0
            ans = 1
            grid[i][j] = 0
            dirs = (-1, 0, 1, 0, -1)
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n:
                    ans += dfs(x, y)
            return ans

        m, n = len(grid), len(grid[0])
        return max(dfs(i, j) for i in range(m) for j in range(n))

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class Solution {
    private int m;
    private int n;
    private int[][] grid;

    public int maxAreaOfIsland(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = Math.max(ans, dfs(i, j));
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        if (grid[i][j] == 0) {
            return 0;
        }
        int ans = 1;
        grid[i][j] = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n) {
                ans += dfs(x, y);
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int dirs[5] = {-1, 0, 1, 0, -1};
        int ans = 0;
        function<int(int, int)> dfs = [&](int i, int j) {
            if (grid[i][j] == 0) {
                return 0;
            }
            int ans = 1;
            grid[i][j] = 0;
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n) {
                    ans += dfs(x, y);
                }
            }
            return ans;
        };
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = max(ans, dfs(i, j));
            }
        }
        return ans;
    }
};

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func maxAreaOfIsland(grid [][]int) (ans int) {
    m, n := len(grid), len(grid[0])
    dirs := [5]int{-1, 0, 1, 0, -1}
    var dfs func(i, j int) int
    dfs = func(i, j int) int {
        if grid[i][j] == 0 {
            return 0
        }
        ans := 1
        grid[i][j] = 0
        for k := 0; k < 4; k++ {
            x, y := i+dirs[k], j+dirs[k+1]
            if x >= 0 && x < m && y >= 0 && y < n {
                ans += dfs(x, y)
            }
        }
        return ans
    }
    for i := range grid {
        for j := range grid[i] {
            ans = max(ans, dfs(i, j))
        }
    }
    return
}

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function maxAreaOfIsland(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const dirs = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number): number => {
        if (grid[i][j] === 0) {
            return 0;
        }
        let ans = 1;
        grid[i][j] = 0;
        for (let k = 0; k < 4; ++k) {
            const [x, y] = [i + dirs[k], j + dirs[k + 1]];
            if (x >= 0 && x < m && y >= 0 && y < n) {
                ans += dfs(x, y);
            }
        }
        return ans;
    };
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            ans = Math.max(ans, dfs(i, j));
        }
    }
    return ans;
}

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impl Solution {
    fn dfs(grid: &mut Vec<Vec<i32>>, i: usize, j: usize) -> i32 {
        if i == grid.len() || j == grid[0].len() || grid[i][j] == 0 {
            return 0;
        }
        grid[i][j] = 0;
        let mut res = 1 + Self::dfs(grid, i + 1, j) + Self::dfs(grid, i, j + 1);
        if i != 0 {
            res += Self::dfs(grid, i - 1, j);
        }
        if j != 0 {
            res += Self::dfs(grid, i, j - 1);
        }
        res
    }

    pub fn max_area_of_island(mut grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let mut res = 0;
        for i in 0..m {
            for j in 0..n {
                res = res.max(Self::dfs(&mut grid, i, j));
            }
        }
        res
    }
}

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