# Description

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

``````Input: k = 3, n = 7
``````

Output:

``````[[1,2,4]]
``````

Example 2:

``````Input: k = 3, n = 9
``````

Output:

``````[[1,2,6], [1,3,5], [2,3,4]]
``````

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<int> nums;
vector<vector<int>> combinationSum3(int k, int n) {
for(int i=1;i<=9;i++)
nums.push_back(i);
int size = nums.size();
if(size <= 0)
return result;
vector<int> oneSet;
oneCombination(oneSet, 0, k, n);
return result;
}
void oneCombination(vector<int> &oneSet, int start, int k, int target) {
if(oneSet.size() > k)
return;
if(target < 0)
return;
else if(target == 0){
for(int i=0;i<result.size();i++){
if(sameSet(result[i], oneSet))
return;
}
if(oneSet.size() == k)
result.push_back(oneSet);
return;
}
for(int i=start;i < nums.size(); i++){
int newTarget = target-nums[i];
oneSet.push_back(nums[i]);
oneCombination(oneSet, i+1, k, 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. Similar to Combination Sum and Combination Sum II, but we need to judge the size of each set in the result.