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>& candidates, int target) { 4 sort(candidates.begin(), candidates.end()); 5 vector<vector<int> > result; 6 vector<int> temp; 7 8 dfs(candidates, target, result, temp, 0); 9 return result; 10 } 11 12 void dfs(vector<int>& candidates, int diff, vector<vector<int> >& result, vector<int>& temp, int start){ 13 if(diff == 0){ 14 result.push_back(temp); 15 return; 16 } 17 18 for(int i = start; i < candidates.size(); i++){ 19 if(diff < candidates[i]) return; 20 21 //only add the first number into the vector if there are many 22 if(i > start && candidates[i] == candidates[i - 1]) continue; 23 24 temp.push_back(candidates[i]); 25 dfs(candidates, diff - candidates[i], result, temp, i + 1); 26 temp.pop_back(); 27 } 28 29 } 30 };