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：

组合问题，递归，1-9 每个数字都分放入和不放入两种情况。得到满足条件的就压入答案。

class Solution {public:    vector
     
      
       > combinationSum3(int k, int n) {        vector
       
         partans;        vector
        
         
          > ans;        combination(1, k, n, partans, ans);        return ans;    }    void combination(int curNum, int k, int sum, vector
          
            &partans, vector
           
            
             > &ans) { if(curNum > 10 || sum < 0 || k < 0) return; //注意这里是 > 10, 否则数字9的答案无法压入 if(sum == 0 && k == 0) { ans.push_back(partans); } else { partans.push_back(curNum); combination(curNum + 1, k - 1, sum - curNum, partans, ans); partans.pop_back(); combination(curNum + 1, k, sum, partans, ans); } }};