class Solution {
public
int numIslands(vector<vector<char>>& grid) {
int n = grid.size();
if(grid.size() <= 0){
return 0;
}
int m = grid[0].size()
int ans = 0;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(grid[i][j] == '0'){
continue;
}
dfs(grid, i, j);
ans += 1;
}
}
return ans;
}
void dfs(vector<vector<char>>& grid, int i, int j)
//i , j 的循环条件,别越界,和判断如果为0,就直接返回即可
if(i > grid.size() - 1 || i < 0 || j < 0 || j > grid[0].size() - 1 || grid[i][j] == '0'){
return;
}
grid[i][j] = '0';
dfs(grid, i + 1, j);
dfs(grid, i, j + 1);
dfs(grid, i - 1, j);
dfs(grid, i, j - 1);
}
};
200. Number of Islands dfs求解
最新推荐文章于 2024-06-25 17:09:24 发布