40.Combination Sum II

[niuworld]

Question:

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

Solution:

class Solution {
public:
    vector<vector<int>>finalresult;
    vector<int>result;

    void helper(vector<int>&can, int tar, vector<int>&re, vector<vector<int>>&fin, int begin)
     {  if( tar == 0){
           fin.push_back(re);
           return;
     }
      for ( int i = begin; ( i <= can.size() - 1 )&&( tar >= can[i]); ++i){
         if( i == begin || can[i] != can[i - 1]){
          re.push_back(can[i]);
          helper(can, tar - can[i], re, fin , i + 1);
          re.pop_back();
          }
      }
     }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        helper(candidates, target, result, finalresult, 0);
        return finalresult;
    }
};

Note: **if( i == begin || can[i] != can[i - 1])**is very important.