216 Combination Sum III-优快云博客

本文探讨了一个经典的组合问题，即寻找所有可能的组合，这些组合由k个数构成，加起来等于给定的数n。讨论了如何使用递归方法找到所有不重复的组合，并给出了详细的代码实现。

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1 题目

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]]

2 尝试解

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> result;
        vector<int> temp;
        find(result,temp,n,k);
        return result;
    }
    void find(vector<vector<int>>&result,vector<int>&temp,int target, int k){
        if(target*target + k*k == 0){ //target=k=0
            result.push_back(temp);
            return;
        }
        if(target*k == 0 || target < 0 || k < 0) return;
        for(int i = (temp.size()==0?1:temp[temp.size()-1] + 1);i <= 9; i++){
            temp.push_back(i);
            find(result,temp,target-i,k-1);
            temp.pop_back();
        }
    }
};

3 标准解

class Solution {
public:
  void combination(vector<vector<int>>& result, vector<int> sol, int k, int n) {
    if (sol.size() == k && n == 0) { result.push_back(sol); return ; }
    if (sol.size() < k) {
      for (int i = sol.empty() ? 1 : sol.back() + 1; i <= 9; ++i) {
        if (n - i < 0) break;
        sol.push_back(i);
        combination(result, sol, k, n - i);
        sol.pop_back();
      }
    }
  }

  vector<vector<int>> combinationSum3(int k, int n) {
    vector<vector<int>> result;
    vector<int> sol;
    combination(result, sol, k, n);
    return result;
  }
};