[LeetCode] Combinations (bfs bad、dfs 递归 accept)

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],
]

方法1：把用queue实现bfs，改为用vector自己实现bfs，没有用额外的内存来存储中间值，但是时间上会Time Limited Exceeded！比如n=13，k=13

class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
        vector<vector<int> > res;
        if(n<1 || k<1)
          return res;
        
        vector<int> v,v0;
        for(int i=1;i<=n;i++){
            v.push_back(i);
            res.push_back(v);
            v.clear();
        }//end for
        
        
        while(!res.empty()){
            v = res[0];
            res.erase(res.begin());
            if(v.size() == k){
                res.push_back(v);
                break;
            }else{
                v0 = v;
                for(int i=1;i<=n;i++){
                    if(find(v.begin(),v.end(),i)==v.end()){
                        v.push_back(i);
                        sort(v.begin(),v.end());
                        if(find(res.begin(),res.end(),v)==res.end())
                           res.push_back(v);
                        v = v0;
                    }
                }//end for
            }
            
        }//end while
        return res;
    }//end func
};

 方法2：用递归（dfs），ACCEPT！

class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
       vector<vector<int> > result;
       vector<int> v;
       combination(n,k,1,result,v);
       return result;  
    }//end func
private:
    void combination(int n,int k,int start,vector<vector<int> > &result,vector<int> l){
        if( k == 0 ){
            result.push_back(l);
            return;
        }

        for(int i =start;i<=n;++i){
            vector<int> a = l;
            a.push_back(i);
            combination(n,k-1,i+1,result,a);
        }
    }//end func
};