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]
class Solution {
public:
vector<int> num;
int target;
vector<vector<int> > result;
vector<vector<int> > combinationSum2(vector<int> &num, int target) {
this->num = num;
this->target = target;
if(num.size()==){
vector<int> tmp;
result.push_back(tmp);
return result;
}
sort(this->num.begin(),this->num.end());
for(int i=;i<num.size();i++){
vector<int> tmp(,this->num[i]);
recursion(i+,tmp,this->num[i]);
}
return result;
}
private:
void recursion(int start,vector<int> tmp,int sum){
if(sum == target && find(result.begin(),result.end(),tmp)==result.end()){
result.push_back(tmp);
return;
}
vector<int> tmp0(tmp);
int sum0 = sum;
for(int i=start;i<num.size();i++){
tmp.push_back(num[i]);
sum += num[i];
if(sum<target)
recursion(i+,tmp,sum);
else if(sum == target){
if(find(result.begin(),result.end(),tmp)==result.end())
result.push_back(tmp);
}
else
break;
tmp = tmp0;
sum = sum0;
}
}//end func
};