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:
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> &num, int target) { vector<vector<int> > ret; if(0 == num.size()) return ret; sort(num.begin(), num.end()); vector<int> digits; digits.push_back(num[0]); vector<int> counts; counts.push_back(1); for(int i = 1; i < num.size(); ++i) { if(digits.back() == num[i]) { ++counts.back(); } else { digits.push_back(num[i]); counts.push_back(1); } } vector<int> result(digits.size()); getAll(digits, counts, result, ret, 0, 0, target); return ret; } void getAll(vector<int>& digits, vector<int>& counts, vector<int>& result, vector<vector<int> >& ret, int curSum, int curIdx, int target) { if(curSum == target) { ret.push_back(vector<int>()); for(int i = 0; i < curIdx; ++i) { for(int j = 0; j < result[i]; ++j) { ret.back().push_back(digits[i]); } } return ; } else if(curSum > target || curIdx == digits.size()) { return ; } else { for(int i = 0; i <= counts[curIdx]; ++i) { result[curIdx] = i; getAll(digits, counts, result, ret, curSum + i*digits[curIdx], curIdx + 1, target); } } } };