[LeetCode 90] child Collection II

Topic Link

【answer】

We want to take a digit in which the enumeration time.
The
1112233
for (int I = Start; I <= n-; I ++)
// where start-1 is the last position taken.
If i> start and num [i] == num [i -1].
So we should no longer take this num [i] a.
Because num certainly would have gone before [i-1] a. At this time, then take a num [i] words. The resulting solution and certainly
11 ***** as the
(i == start where it is obtained it can take 111 ****)

[Code]

class Solution {
public:
    vector<vector<int>> ans;

void dfs(vector<int> &v,int start,vector<int> &temp){
    ans.push_back(temp);
    for (int i = start;i<(int)v.size();i++){
        if (i>start && v[i]==v[i-1]) continue;
        temp.push_back(v[i]);
        dfs(v,i+1,temp);
        temp.pop_back();
    }
}

vector<vector<int>> subsetsWithDup(vector<int> v){
    sort(v.begin(),v.end());
    ans.clear();
    vector<int> t;t.clear();
    dfs(v,0,t);
    return ans;
}

};