forked from havanagrawal/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_695.java
66 lines (54 loc) · 1.83 KB
/
_695.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
package com.fishercoder.solutions;
/**
* 695. Max Area of Island
*
* 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.
*/
public class _695 {
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;
}
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;
}
}