Combination Sum III

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]]

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

参考Combination Sum II,把去重条件去掉(1-9本身就不包含重复元素),仅返回包含k个元素的vector

class Solution {
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<int> num();
for(int i = ; i < ; i ++)
num[i] = i+;
return combinationSum2(num, n, k);
}
vector<vector<int> > combinationSum2(vector<int> &num, int target, int k) {
sort(num.begin(), num.end());
vector<vector<int> > ret;
vector<int> cur;
Helper(ret, cur, num, target, , k);
return ret;
}
void Helper(vector<vector<int> > &ret, vector<int> cur, vector<int> &num, int target, int position, int k)
{
if(target == )
{
if(cur.size() == k)
ret.push_back(cur);
else
;
}
else
{
for(int i = position; i < num.size() && num[i] <= target; i ++)
{
cur.push_back(num[i]);
Helper(ret, cur, num, target-num[i], i+, k);
cur.pop_back();
}
}
}
};

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