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]]

解答：

1.我的方法，回溯

//
//  main.cpp
//  Combination Sum III
//
//  Created by zjl on 16/4/29.
//  Copyright © 2016年 zjl. All rights reserved.
//

#include <iostream>
#include <vector>
using namespace std;

void combination(vector<vector<int>>& res,vector<int> list,vector<int>& vec,int begin,int k,int n){
    if(k==0 && n==0){
        res.push_back(vec);
        return;
    }
    for(int i = begin; i < list.size(); i++){
        if(n >= list[i]){
            vec.push_back(list[i]);
            combination(res,list,vec,i+1,k-1,n-list[i]);
            vec.pop_back();
        }else return;
    }
}

vector<vector<int>> combinationSum3(int k, int n) {
    vector<vector<int>> res;
    vector<int> vec;
    vector<int> list;
    int begin = 0;
    if(k >= n)
        return res;
    for(int i = 0; i < n; i++)
        list.push_back(i+1);
    combination(res,list,vec,begin,k,n);
    return res;
}
void print_vec(vector<vector<int>> v){
    for(int i = 0; i < v.size(); i++){
        for(int j = 0; j < v[i].size(); j++){
            cout<<v[i][j]<<" ";
        }
        cout<<endl;
    }
}

int main(int argc, const char * argv[]) {
    int k = 3;
    int n = 9;
    vector<vector<int>>res = combinationSum3(k, n);
    print_vec(res);
    return 0;
}

输出：

1 2 6 

1 3 5 

2 3 4

注：这里

combination(res,list,vec,i+1,k-1,n-list[i]);

和

combination(res,list,vec,begin+1,k-1,n-list[i]);

输出的答案是不一样的。

以 i+1 输出的是从i以后的遍历得到的结果，就如以上答案所示。若以begin+1传递，则是下次遍历也从begin开始。得到的答案为以下：

1 2 6 

1 3 5 

1 4 4 

1 5 3 

2 2 5 

2 3 4 

2 4 3 

3 2 4 

3 3 3 

4 2 3