39-Combination Sum
100-Same Tree | Links:
题意:
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: []
Example 4:
- Input: candidates = [1], target = 1
- Output: [[1]]
Example 5:
- Input: candidates = [1], target = 2
- Output: [[1,1]]
Constraints:
- 1 <= candidates.length <= 30
- 1 <= candidates[i] <= 200
- All elements of candidates are distinct.
- 1 <= target <= 500
思路
- 回溯法
- 结果需要满足sum条件:用target
- 结果中不能有重复组合:用index控制,但可以有重复元素,index不增加
class Solution(object):
def combinationSum(self, candidates, target):
"""
:type candidates: List[int]
:type target: int
:rtype: List[List[int]]
"""
res = []
if not candidates or target == 0: return res
def backtrack(res, temp, nums, target, start):
if target < 0: return
if target == 0:
res.append(temp[:])
return
for i in range(start, len(nums)):
temp.append(nums[i])
backtrack(res, temp, nums, target - nums[i], i)
temp.pop()
backtrack(res, [], candidates, target,0)
return res
分析
- Time: O(\(2^n\))
- Space:O(n)