題目
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
For example,
If n = 4 and k = 2, a solution is:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
頻度： 4
解題之法
class Solution {
public:
vector<vector<int> > combine(int n, int k) {
vector<vector<int> >res;
if(n<k)return res;
vector<int> temp(0,k);
combine(res,temp,0,0,n,k);
return res;
}
void combine(vector<vector<int> > &res,vector<int> &temp,int start,int num,int n ,int k){
if(num==k){
res.push_back(temp);
return;
}
for(int i = start;i<n;i++){
temp.push_back(i+1);
combine(res,temp,i+1,num+1,n,k);
temp.pop_back();
}
}
};
分析
DFS
The idea is using backtracking ,every time I push a number into vector,then I push a bigger one into it;
then i pop the latest one,and push a another bigger one...
and if I has push k number into vector,I push this into result;
