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]

class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int> > result;
vector<int> temp;

dfs(candidates, target, result, temp, 0);
return result;
}

void dfs(vector<int>& candidates, int diff, vector<vector<int> >& result, vector<int>& temp, int start){
if(diff == 0){
result.push_back(temp);
return;
}

for(int i = start; i < candidates.size(); i++){
if(diff < candidates[i]) return;

//only add the first number into the vector if there are many
if(i > start && candidates[i] == candidates[i - 1]) continue;

temp.push_back(candidates[i]);
dfs(candidates, diff - candidates[i], result, temp, i + 1);
temp.pop_back();
}

}
};