1905_CountSubIslands - a920604a/leetcode GitHub Wiki
title: 1905. Count Sub Islands
tags:
- backtracking
- bfs
- Union Find
categories: leetcode
class Solution {
public:
void dfs(vector<vector<int>>& grid, int i, int j){
int n = grid.size(), m = grid[0].size();
if(i< 0 || j<0 || i>n-1 || j>m-1 ||grid[i][j] == 0 ) return;
grid[i][j] = 0;
dfs(grid,i-1,j);
dfs(grid, i+1,j);
dfs(grid, i,j-1);
dfs(grid, i,j+1);
}
int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
int n = grid1.size(), m = grid1[0].size();
for(int i=0;i<n;++i){
for(int j=0;j<m;++j){
// 如果grid2 的島嶼但不是grid1的島嶼,淹沒
// 這個島於肯定不是子島嶼,淹沒
if(grid1[i][j] == 0 && grid2[i][j] == 1) dfs(grid2, i, j);
}
}
int count = 0;
for(int i =0;i<n;i++){
for(int j=0;j<m;++j){
if(grid2[i][j] == 1){
count++;
dfs(grid2, i, j);
}
}
}
return count;
}
};
- time complexity
O(n*m)
- space complexity
O(n*m)