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.

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.

解题思路：

深搜？因为在9之内所以时间应该不会很高。

 

class Solution {
    private:

     vector<vector<int>> ret;
    vector<int> store;

    void deep(int &k,int &n,int cur_num,int count,int sum){
        count++;
        sum+=cur_num;
        store.push_back(cur_num);
        if(count == k){
            if(n==sum) {ret.push_back(store);return;}
        }
        else{
            for(int j=cur_num+1;j<=9;j++){
                deep(k,n,j,count,sum);
                store.pop_back();
            }
        }
    }
public:
    vector<vector<int>> combinationSum3(int k, int n) {


        for(int i=1;i<=9;i++){
            if((9-i+1)>=k)
            deep(k,n,i,0,0);
            store.clear();

        }
        return ret;
    }
};