Combination Sum III

Description

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

• Only numbers 1 through 9 are used.
• Each number is used at most once.

Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

Example 1:

Input: k = 3, n = 7 Output: 1,2,4 Explanation: 1 + 2 + 4 = 7 There are no other valid combinations.

Example 2:

Input: k = 3, n = 9 Output: [[1,2,6],[1,3,5],[2,3,4]] Explanation: 1 + 2 + 6 = 9 1 + 3 + 5 = 9 2 + 3 + 4 = 9 There are no other valid combinations.

Example 3:

Input: k = 4, n = 1 Output: [] Explanation: There are no valid combinations. Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.

Constraints:

• 2 <= k <= 9
• 1 <= n <= 60

Code

結合 Combination Sum II 以及 Combinations 的概念。

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> answer;
        vector<int> curr;
        vector<int> candidates = {1, 2, 3, 4, 5, 6, 7, 8, 9};
        helper(answer, candidates, curr, 0, 0, n, k);
        return answer;
    }
 
    void helper(vector<vector<int>> &answer, vector<int> &candidates, vector<int> curr, int currSum, int start, int target, int k) {
 
        if(currSum > target || curr.size() > k) return;
        if(currSum == target && curr.size() == k) {
            answer.push_back(curr);
            return;
        }
 
        for(int i = start; i < candidates.size(); i++) {
 
            currSum += candidates[i];
            curr.push_back(candidates[i]);
            // set start to i + 1, since elements cannot be reuse
            helper(answer, candidates, curr, currSum, i + 1, target, k);
            currSum -= candidates[i];
            curr.pop_back();
        }
    }
 
};

Source

• Combination Sum III - LeetCode