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]
思路：和Combination Sum这道题基本思想是相同的。需要注意的不同点主要有两个：一个是因为每个数只能用一次，因此在枚举完后继续递归时要从下一位开始枚举，而不再是当前位置；另一个还是因为每个数只能用一次，因此要注意原数组中可能有重复数字的情况，因此这里在枚举过程中如果遇见了相同的数，则只枚举一次，然后跳过剩下的重复数字。
1 class Solution { 2 public: 3 void assist(vector<int>& candidates, vector<vector<int> >& res, vector<int>& cur, int target, int st) 4 { 5 if (target == 0) 6 { 7 res.push_back(cur); 8 return; 9 } 10 if (st == candidates.size()) return; 11 for (int i = st, n = candidates.size(); i < n && candidates[i] <= target; i++) 12 { 13 cur.push_back(candidates[i]); 14 assist(candidates, res, cur, target - candidates[i], i + 1); 15 cur.pop_back(); 16 while (i < n && candidates[i] == candidates[i + 1]) 17 i++; 18 } 19 } 20 vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { 21 vector<vector<int> > res; 22 vector<int> cur; 23 sort(candidates.begin(), candidates.end(), less<int>()); 24 assist(candidates, res, cur, target, 0); 25 return res; 26 } 27 };