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
A solution set is:
» Solve this problem10,1,2,7,6,1,5
and target 8
, A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
[Thoughts]
Very similar with previous "Combination Sum". The only difference is marked as red in Code part. Need to handle the index and skip duplicate candidate.
[Code]
1: vector<vector<int> > combinationSum2(vector<int> &num, int target) { 2: // Start typing your C/C++ solution below 3: // DO NOT write int main() function 4: // Start typing your C/C++ solution below 5: // DO NOT write int main() function 6: vector<vector<int> > result; 7: vector<int> solution; 8: int sum=0; 9: std::sort(num.begin(), num.end()); 10: GetCombinations(num,sum, 0, target, solution, result); 11: return result; 12: } 13: void GetCombinations( 14: vector<int>& candidates, 15: int& sum, 16: int level, 17: int target, 18: vector<int>& solution, 19: vector<vector<int> >& result) 20: { 21: if(sum > target) return; 22: if(sum == target) 23: { 24: result.push_back(solution); 25: return; 26: } 27: for(int i =level; i< candidates.size(); i++) 28: { 29: sum+=candidates[i]; 30: solution.push_back(candidates[i]); 31: GetCombinations(candidates, sum,
i+1
, target, solution, result); 32: solution.pop_back(); 33: sum-=candidates[i]; 34:
while(i<candidates.size()-1 && candidates[i] == candidates[i+1]) i++;
35: } 36: }
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