1、题目来源
2、题目描述
给你一个由 '1'
(陆地)和 '0'
(水)组成的的二维网格,请你计算网格中岛屿的数量。
岛屿总是被水包围,并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
此外,你可以假设该网格的四条边均被水包围。
示例 1:
输入:grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出:1
示例 2:
输入:grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出:3
提示:
m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j]
的值为'0'
或'1'
3、题解分享
java
class Solution {
public int numIslands(char[][] grid) {
// 思路:深度遍历 + 修改遍历过的位置
int n = grid.length;
int m = grid[0].length;
int ans = 0;
for(int i=0;i<n;++i){
for(int j = 0;j<m;++j){
if(grid[i][j] == '1'){//每遇到一块岛屿
++ans;
dfs(grid,n,m,i,j);
}
}
}
return ans;
}
public void dfs(char[][] grid,int n,int m,int row,int col){
if(row <0 || row >= n || col < 0 || col >=m){
return;
}
if(grid[row][col] == '0'){
return;
}
grid[row][col] = '0';
dfs(grid,n,m,row-1,col);
dfs(grid,n,m,row+1,col);
dfs(grid,n,m,row,col-1);
dfs(grid,n,m,row,col+1);
}
}