Leetcode -- Combination sum II

Idea: similar to the previous question, but an index array is added to record the stored element index in the result,

Used to determine whether the current element is the same as the previous one and whether the previous one has been used.

It is mainly used to solve the problem of repeated solutions.

 1 class Solution { 2 public: 3     vector<vector<int> >ans; 4     vector<vector<int> > combinationSum2(vector<int> &num, int target) { 5         if(num.size() == 0) 6             return ans; 7         sort(num.begin(),num.end()); 8         vector<int> res; 9         vector<int> index;10         calSum(num,target,0,res,index);11         return ans;12     }13     void calSum(vector<int> &num,int target,int s,vector<int> &res,vector<int> &index)14     {15         if(target < 0)16             return;17         if(target == 0)18         {19             ans.push_back(res);20             return;21         }22         int i;23         for(i = s ; i < num.size() ; ++i)24         {25             if(i >0 && index[index.size()-1] != i-1 && num[i] == num[i-1])26                 continue;27             index.push_back(i);28             res.push_back(num[i]);29             calSum(num,target-num[i],i+1,res,index);30             res.pop_back();31             index.pop_back();32         }33         return;34     }35 };