216. 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.

Note:

• All numbers will be positive integers.
• The solution set must not contain duplicate combinations.

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]]



与上面的题不同，此题的候选集变成1，2,3,4,5,6,7，8，9。另外对output的长度有限制。所以在搜索的时候我们要记录当前长度
当output的长度不符合要求的时候，及时return

class Solution {
public:
    vector<vector<int>> res;
    void backtrack(vector<int>& path,int k ,int n,int sum, int start) {
        if (path.size() == k && sum == n) {
            res.push_back(path);
            return;
        }
        for(int i = start; i <= 9 ; ++i) {
            if(sum > n) {
                continue;
            }
            path.push_back(i);
            backtrack(path,k,n,sum+i,i+1);
            path.pop_back();
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<int> path;
        backtrack(path,k,n,0,1);
        return res;
    }
};

 

 1 class Solution {
 2 public:
 3     vector<vector<int>> finalres ;
 4     vector<vector<int>> combinationSum3(int k, int target) {
 5         vector<int> curres;
 6         help(0,curres,1,k,target);
 7         return finalres;
 8         
 9     }
10     void help(int cursum,vector<int>& curres,int index,int k,int target){
11         if(cursum==target&& k==0)
12             finalres.push_back(curres);
13         if(cursum>target||k<0)
14             return;
15         for(int i = index;i<=9;i++){
16             curres.push_back(i);
17             help(cursum,curres,i+1,k-1,target-i);
18             curres.pop_back();
19 
20         }
21     }
22 };