# Description

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]

The original problem is here.

The original code is here.

# My Solution

I solve this problem in C++, as below:

``````/*
*Combination Sum
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<math.h>
#include<vector>
#include<stdlib.h>
using namespace std;

class Solution {
public:
vector<vector<int>> result;
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
int size = candidates.size();
if(size <= 0)
return result;
vector<int> oneSet;
sort(candidates.begin(),candidates.end());
oneCombination(candidates, oneSet, 0, target);
return result;
}
void oneCombination(vector<int>&candidates, vector<int> &oneSet, int start, int target) {
if(target < 0)
return;
else if(target == 0){
for(int i=0;i<result.size();i++){
if(sameSet(result[i], oneSet))
return;
}
result.push_back(oneSet);
return;
}
for(int i=start;i < candidates.size(); i++){
int newTarget = target-candidates[i];
oneSet.push_back(candidates[i]);
oneCombination(candidates, oneSet, i+1, newTarget);
oneSet.pop_back();
}
return;
}
bool sameSet(vector<int> set1, vector<int> set2) {
int size1 = set1.size();
int size2 = set2.size();
if(size1 != size2)
return false;
for(int i=0;i<size1;i++) {
if(set1[i] != set2[i])
return false;
}
return true;
}
};
``````

# Note

To solve the problem, recursion is used. We also need to judge whether the reslt has duplicate combinations.