LeetCode 216 Combination Sum III

LeetCode

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

题目大意：

数字1-9的组合个数为k的组合和等于k，每个元素最多用一次

解题思路：

用组合模板，先排序

解题步骤：

N/A

注意事项：

1. Leetcode 40和77的结合。个数和target都要达到。用if k == 0 and target == 0

Python代码：

def combinationSum3(self, k: int, n: int) -> List[List[int]]:
	nums = [_ for _ in range(1, 10)]
	res = []
	self.dfs(nums, 0, k, n, [], res)
	return res

def dfs(self, nums, start, k, target, path, res):
	if k == 0 and target == 0:
		res.append(list(path))
		return
	if k == 0:
		return
	for i in range(start, len(nums)):
		path.append(nums[i])
		self.dfs(nums, i + 1, k - 1, target - nums[i], path, res)
		path.pop()

算法分析：

时间复杂度为O(2k)，空间复杂度O(n)