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- Combination Sum
- 难度：Medium｜ 中等
- 相关知识点：Backtracking
- 题目链接：https://leetcode.com/problems/combination-sum/
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.
The test cases are generated such 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
2 <= candidates[i] <= 40
All elements of candidates are distinct.
1 <= target <= 40
Solution 1:
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
l = len(candidates)
def backtrack(start, target, ans):
if target == 0:
result.append(ans[:])
return
for index in range(start, len(candidates)):
c = candidates[index]
if target - c >= 0:
ans.append(c)
backtrack(index, target-c, ans)
ans.pop()
result = []
candidates.sort()
backtrack(0, target, [])
return result