the original solution:

```
public int numIslands(char[][] grid) {
int result = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == '1') {
result++;
dfs(grid, i, j);
}
}
}
return result;
}
private void dfs(char[][] grid, int i, int j) {
if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] != '1') {
return;
}
grid[i][j] = '2';
dfs(grid, i, j + 1);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i - 1, j);
}
```

I met a variant of this question in the interview of Baidu, the problem is to find the maximum area of the island, which changes the original code a little bit:

```
private static int num;
public static int bigHomes(int[][] grid) {
int result = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
num = 0;
dfs(grid, i, j);
result = Math.max(result, num);
}
}
}
return result;
}
private static void dfs(int[][] grid, int i, int j) {
if (i < 0 || i <= grid.length || j < 0 || j >= grid[0].length || grid[i][j] != 1) {
return;
}
num++;
grid[i][j] = 2;
dfs(grid, i, j + 1);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i - 1, j);
}
```