LeetCode [216] Combination Sum III

 216. Combination Sum III

Medium

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. [1,2,1] is not valid because 1 is used twice.

Example 4:

Input: k = 3, n = 2
Output: []
Explanation: There are no valid combinations.

Example 5:

Input: k = 9, n = 45
Output: [[1,2,3,4,5,6,7,8,9]]
Explanation:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
​​​​​​​There are no other valid combinations.

 

Constraints:

• 2 <= k <= 9
• 1 <= n <= 60

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class Solution { public: vector<vector<int>> combinationSum3(int k, int n) { vector<vector<int>> res; vector<int> cur; bt(k, n, res, cur, 1, 0); return res; } void bt(int k, int n, vector<vector<int>> &res, vector<int> cur, int pos, int sum){ if(sum==n && cur.size()==k){ res.push_back(cur); }else if(sum<n && pos<=9 && cur.size()<k){ for(int i=pos; i<=9; ++i){ if(sum+i>n) return; cur.push_back(i); bt(k, n, res, cur, i+1, sum+i); cur.pop_back(); } } } };

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class Solution { List<List<Integer>> lists = new ArrayList<>(); public List<List<Integer>> combinationSum3(int k, int n) { helper(0, new ArrayList<>(), k, n, 1); return lists; } void helper(int sum, List<Integer> list, int k, int n, int p){ if(sum>=n){ if(sum==n && list.size()==k){ lists.add(list); } }else if(p<=9 && sum+p<=n && list.size()<k){ helper(sum, list, k, n, p+1); List<Integer> newList = new ArrayList<Integer>(list); newList.add(p); helper(sum+p, newList, k, n, p+1); } } }