LeetCode 039 Combination Sum

LeetCode

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1
Output: []

Constraints:

1 <= candidates.length <= 30 1 <= candidates[i] <= 200
All elements of candidates are distinct. 1 <= target <= 500

题目大意：

求组合和等于目标。元素可以复用

解题思路：

用组合模板，先排序

解题步骤：

N/A

注意事项：

1. 用标准组合模板dfs(self, candidates, start, target, path, res)，元素可以复用，所以下一轮递归从i开始
2. Python中path.pop()没有参数

Python代码：

def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
	candidates.sort()
	res = []
	self.dfs(candidates, 0, target, [], res)
	return res

def dfs(self, candidates, start, target, path, res): # [1, 2], 0, 0, [1, 1], [1, 1]
	if target < 0:
		return
	if target == 0:
		res.append(list(path))
		return
	for i in range(start, len(candidates)): # [2]
		path.append(candidates[i]) # [1,1]
		self.dfs(candidates, i, target - candidates[i], path, res)
		path.pop()

算法分析：

时间复杂度为O(2n)，空间复杂度O(n)