Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.
An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
Input: grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
Output: 1
Example 2:
Input: grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
Output: 3
Constraints:
- m == grid.length
- n == grid[i].length
- 1 <= m, n <= 300
- grid[i][j] is '0' or '1'.
class Solution {
private final char LAND = '1';
private final char WATER = '0';
private char[][] arr;
public int numIslands(char[][] grid) {
arr = grid;
int nums = 0;
for(int i=0;i<grid.length;i++){
for(int j=0;j<grid[0].length;j++){
if(grid[i][j] == LAND){
nums++;
dfs(i,j);
}
}
}
return nums;
}
private void dfs(int rows, int cols){
// check if current rows is out of bound
if(rows<0||rows>arr.length-1) return;
// check if current cols is out of bound
if(cols<0||cols>arr[0].length-1) return;
// WATER or LAND
if(arr[rows][cols]==WATER) return;
arr[rows][cols] = WATER;
dfs(rows+1, cols);
dfs(rows-1, cols);
dfs(rows, cols+1);
dfs(rows, cols-1);
}
}