Combination Sum II

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

在上一题基础上增加判断，如果一个数不选择，那么之后所有相同的数字也都不选择。这组题类似subsets那一组。

 1     void search(vector<vector<int> > &result, vector<int> &tmp, vector<int> &candidates, int index, int target){
 2         int n = candidates.size();
 3         if(target == 0){
 4             if(((index < n && (!tmp.empty() && candidates[index] != *(tmp.end()-1)))) || index >= n)
 5                 result.push_back(tmp);
 6             return;
 7         }
 8         if(index >= n)
 9             return;
10         if(target < candidates[index])
11             return;
12         if(tmp.empty() || (!tmp.empty() && *(tmp.end()-1) != candidates[index]))
13             search(result, tmp, candidates, index+1, target);
14         tmp.push_back(candidates[index]);
15         search(result, tmp, candidates, index+1, target-candidates[index]);
16         tmp.erase(tmp.end()-1);
17     }
18     vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
19         // Start typing your C/C++ solution below
20         // DO NOT write int main() function
21         vector<vector<int> > result;
22         vector<int> tmp;
23         sort(candidates.begin(), candidates.end());
24         search(result, tmp, candidates, 0, target);
25         return result;
26     }