695. Max Area of Island
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 either0
or1
.
class Solution { public int maxAreaOfIsland(int[][] grid) { int res = 0; for(int i = 0; i < grid.length; i++) { for(int j = 0; j < grid[0].length; j++) { if(grid[i][j] == 1) { res = Math.max(res, dfs(i, j, grid)); } } } return res; } public int dfs(int i, int j, int[][] grid) { if(i >= grid.length || i < 0 || j < 0 || j >= grid[0].length || grid[i][j] == 0) return 0; grid[i][j] = 0; return 1 + dfs(i, j - 1, grid)+ dfs(i-1, j, grid)+ dfs(i+1, j, grid)+ dfs(i, j+1, grid); } }
similar to #200.