Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
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 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
使用递归的方法实现：
#include<iostream> #include<vector> #include<algorithm> using namespace std; class Solution { public: vector<vector<int> > combinationSum(vector<int> &candidates, int target) { if(candidates.empty()) return vector<vector<int> >(); sort(candidates.begin(),candidates.end()); vector<vector<int> > ret; vector<int> path; combination(candidates,0,target,ret,path); return ret; } void combination(vector<int> &candidates,int start,int target,vector<vector<int> > &ret,vector<int> &path) { if(target<0) return; if(target==0) { ret.push_back(path); return; } int i; for(i=start;i<(int)candidates.size();i++) { path.push_back(candidates[i]); combination(candidates,i,target-candidates[i],ret,path); path.pop_back(); } } }; int main() { Solution s; vector<int> vec={2,3,6,7}; vector<vector<int> > result=s.combinationSum(vec,7); for(auto a:result) { for(auto v:a) cout<<v<<" "; cout<<endl; } }
运行结果：
