Leetcode 0216: Combination Sum III

题目描述：

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

Time complexity: $O({\mathrm{C}}_9^k$ * k)
Space Complexity: $O(k)$
DFS Backtracking：
暴力搜索出所有从9个数中选 k 个的方案，记录所有和等于 n 的方案。
为了避免重复计数, 如果上一个选的数是 x，那我们从 x+1 开始枚举当前数。

时间复杂度分析：从9个数中选 k 个总共有 ${\mathrm{C}}_9^k$ 个方案，将每个方案记录下来需要 O(k) 的时间，所以时间复杂度是 O(${\mathrm{C}}_9^k$ * k)。

class Solution {
    
    
    List<List<Integer>> res = new ArrayList<>();
    public List<List<Integer>> combinationSum3(int k, int n) {
    
    
        dfs(1, new ArrayList<Integer>(), k, n);
        return res;
    }
    
    private void dfs(int next, List<Integer> path, int k, int remain){
    
    
        if(k == path.size() && remain == 0){
    
    
            res.add(new ArrayList<Integer>(path));
            return;
        }
        for(int i = next; i <=9; i++){
    
    
            path.add(i);
            dfs(i+1, path, k, remain-i);
            path.remove(path.size()-1);
        }
    }
}