LeetCode-in-Java

695. Max Area of Island

Medium

You are given an m x n binary matrix grid. 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.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

Example 1:

Input: 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]]

Output: 6

Explanation: The answer is not 11, because the island must be connected 4-directionally.

Example 2:

Input: grid = [[0,0,0,0,0,0,0,0]]

Output: 0

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• grid[i][j] is either 0 or 1.

Solution

public class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }
        int m = grid.length;
        int n = grid[0].length;
        int max = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    int area = dfs(grid, i, j, m, n, 0);
                    max = Math.max(area, max);
                }
            }
        }
        return max;
    }

    private int dfs(int[][] grid, int i, int j, int m, int n, int area) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
            return area;
        }
        grid[i][j] = 0;
        area++;
        area = dfs(grid, i + 1, j, m, n, area);
        area = dfs(grid, i, j + 1, m, n, area);
        area = dfs(grid, i - 1, j, m, n, area);
        area = dfs(grid, i, j - 1, m, n, area);
        return area;
    }
}

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