Medium - Array
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
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
answer = []
def DFS(current, startIdx, path): # path -> 원소를 뺀 기록. 누적 경로
if current < 0: # 뺐을 때 0이하로 떨어져서 필요 없는 탐색 경로
return
elif current == 0: # 원소값을 뻈을 때 0이 되서 찾는 타겟값이 되는 경로
answer.append(path)
return
for i in range(startIdx, len(candidates)):
DFS(current - candidates[i], i, path + [candidates[i]])
# DFS로 중복으로 원소값을 뺀 경우를 i로 다 탐색
DFS(target, 0, [])
return answer
참고)
https://velog.io/@ahn16/Leetcode-39-Python-Combination-Sum
