Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums toT.
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]
暴力搜尋
class Solution { public: vector<vector<int> > res; vector<int> path; vector<int> candidates; int target; void solve(int start, int sum){ if(sum > target) return; if(sum == target){ if(find(res.begin(),res.end(),path)==res.end()) res.push_back(path); return; } for(int i = start; i < candidates.size(); ++ i){ path.push_back(candidates[i]); solve(i+1,sum+candidates[i]); path.pop_back(); } } vector<vector<int> > combinationSum2(vector<int> &candidates, int target) { this->candidates = candidates; this->target = target; sort(this->candidates.begin(),this->candidates.end()); solve(0,0); return res; } };