216. Combination Sum III-优快云博客

本文探讨了如何找到所有可能的组合，这些组合由k个数构成且总和为n，仅使用1到9之间的数字，并确保每个组合都是唯一的数字集。通过递归回溯的方法，提供了一个C++实现的解决方案，有效地解决了这一问题，避免了重复的组合。

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]]

Approach #1: BackTracking. [C++]

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<int> sol;
        vector<vector<int>> ans;
        helper(k, n, sol, ans);
        return ans;
    }
private:
    void helper(const int k, int n, vector<int> v, vector<vector<int>>& ans) {
        if (v.size() == k && n == 0) {
            ans.push_back(v);
            return;
        }
        if (v.size() < k) {
            for (int i = v.empty()?1:v.back()+1; i <= 9; ++i) {
                if (n - i < 0) break;
                v.push_back(i);
                helper(k, n-i, v, ans);
                // this is I don't care in the first time.
                v.pop_back();
            }   
        }
    }
};

转载于:https://www.cnblogs.com/ruruozhenhao/p/10347250.html