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]]

[思路]

因为k是不定的, 所以无法用LOOP. 递归是此类题的常用解法. 为了方便, 多用一个variable: sum保存当前cur中所有数的和. 注意要新建一个list 放入结果中, 否则放入的reference 会指向原来的不断变化的list,    res.add(new ArrayList(cur));

[CODE]

class Solution {
public:
    void visit(int pos, int k, int n, int &sum,  vector<int> &buf, vector<vector<int> > &result)
    {
        if (k == 0)
        {
            if (sum == n)
            {
                result.push_back(buf);
            }
            return;
        }
        for (int i = pos; i <= 9; i++)
        {   
            if (sum + i > n)
            {
                break;
            }
            buf.push_back(i);
            sum += i;
            visit(i+1, k-1, n, sum, buf, result);
            buf.pop_back();
            sum -= i;
        }
    }
    
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int> > result;
        vector<int> buf;
        int sum = 0;
        visit(1, k, n, sum, buf, result);
        return result;
    }
};