README_EN.md

695. Max Area of Island

中文文档

Description

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[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]]

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.

Solutions

Python3

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        def dfs(grid, i, j, m, n):
            if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] == 0:
                return 0
            grid[i][j] = 0
            res = 1
            for x, y in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
                res += dfs(grid, i + x, j + y, m, n)
            return res
        
        m, n = len(grid), len(grid[0])
        res = 0
        for i in range(m):
            for j in range(n):
                t = dfs(grid, i, j, m, n)
                res = max(res, t)
        return res

Java

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

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

    private int dfs(int[][] grid, int i, int j, int m, int n) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
            return 0;
        }
        grid[i][j] = 0;
        int res = 1;
        for (int[] direction : directions) {
            res += dfs(grid, i + direction[0], j + direction[1], m, n);
        }
        return res;
    }
}

TypeScript

function maxAreaOfIsland(grid: number[][]): number {
    let m = grid.length, n = grid[0].length;
    let res = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] == 1) {
                res = Math.max(dfs(grid, i, j), res);
            }
        }
    }
    return res;
};

function dfs(grid: number[][], i: number, j: number): number {
    let m = grid.length, n = grid[0].length;
    if (i < 0 || i > m - 1 || j < 0 || j > n - 1 || grid[i][j] == 0) {
        return 0;
    }
    grid[i][j] = 0;
    let res = 1;
    for (let [dx, dy] of [[0, 1], [0, -1], [1, 0], [-1, 0]]) {
        res += dfs(grid, i + dx, j + dy);
    }
    return res;
}

C++

class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int res = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int t = dfs(grid, i, j, m, n);
                res = max(res, t);
            }
        }
        return res;
    }
private:
    vector<vector<int>> directions = {{0, 1}, {0, - 1}, {1, 0}, {-1, 0}};

    int dfs(vector<vector<int>>& grid, int i, int j, int m, int n) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
            return 0;
        }
        grid[i][j] = 0;
        int res = 1;
        for (auto direction : directions) {
            res += dfs(grid, i + direction[0], j + direction[1], m, n);
        }
        return res;
    }

};

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