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:
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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)
}