Leetcode: Combination Sum III-优快云博客

本文链接：https://blog.youkuaiyun.com/u012309913/article/details/54912103

本文介绍了一个算法问题：寻找所有可能的组合，使得k个数的和等于目标数n，且只使用1到9的数字，每组组合中的数字不重复。文章通过递归回溯的方法实现了这一功能，并给出了具体的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Subscribe to see which companies asked this question.

class Solution {
private:
   int index_count;
   vector<vector<int> > results;
public:
    void backtrace(int target, int sum, vector<int>& index, int id, int n, int k)
    {
        if (sum > target || n > k) {
            return;
        }
        else if (sum == target && n == k) {
            vector<int> result;
            for(int i = 1; i <= n; ++i)
            {
                result.push_back(index[i]);
            }
            results.push_back(result);
            return;
        }
        
        for (int i = id + 1; i < 10; ++i)
        {
            index[n+1] = i;
            backtrace(target, sum + i, index, i, n+1, k);
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        if (k * 9 < n) {
            return results;
        }
        
        vector<int> index = vector<int>(10, 0);
        for (int i = 0; i < 10; ++i) {
            index[i] = i;
        }
        results.clear();
        backtrace(n, 0, index, 0, 0, k);
        
        return results;
        
        
    }
};