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.

解題

func maxAreaOfIsland(grid [][]int) int {
    max := 0

    for i := 0; i < len(grid); i++ {
        for j := 0; j < len(grid[0]); j++ {
            area := dfs(grid, i, j)
            if area > max {
                max = area
            }
        }
    }

    return max
}

func dfs(grid [][]int, x int, y int) int {
    if x < 0 || x > len(grid) - 1 || y < 0 || y > len(grid[0]) - 1 || grid[x][y] == 0 {
        return 0
    }

    grid[x][y] = 0
    
    return 1 + dfs(grid, x + 1, y) + dfs(grid, x - 1, y) + dfs(grid, x , y - 1) + dfs(grid, x, y + 1)
}