LeetCode-40-Combination Sum ||

算法描述：

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

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

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

解题思路：求所有可能性的题目，首先想到回溯法。这道题目中的重点是去重。

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> results;
        vector<int> temp;
        combine(candidates, temp, target, results, 0);
        return results;
    }
    
    void combine(vector<int>& candidates, vector<int>& temp, int target, vector<vector<int>>& results, int begin){
        if(target ==0){
            results.push_back(temp);
            return;
        }
        for(int i=begin; i < candidates.size() && target >= candidates[i]; i++){
            if(i > begin && candidates[i]==candidates[i-1]) continue;
            temp.push_back(candidates[i]);
            combine(candidates,temp,target-candidates[i],results,i+1);
            temp.pop_back();
        }
    }