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]
1 class Solution {
2 public:
3 vector<vector<int> > combinationSum2(vector<int> &num, int target) {
4 sort(num.begin(), num.end());
5 vector<vector<int> > res;
6 vector<int> com;
7 combinationSum2(num, target, res, com, 0);
8 return res;
9 }
10
11 void combinationSum2(vector<int>& num, int target, vector<vector<int> > &res, vector<int> &com, int start)
12 {
13 if(target == 0) {
14 res.push_back(com);
15 return;
16 }
17 for(int i = start; i < num.size() && num[i] <= target; i++) {
18 if(i > start && num[i] == num[i-1]) continue; // !duplicate cases
19 com.push_back(num[i]);
20 combinationSum2(num, target-num[i], res, com, i+1);
21 com.pop_back();
22 }
23 }
24 };
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原文：http://www.cnblogs.com/zhengjiankang/p/3646360.html