39. Combination Sum

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]]

找數字組合。

1. 目標比列表中最小值還小就返回空
   target < min(candidates)
2. 下次迭代用target-candidate當新target
   self.combinationSum(candidates[i:],target-candidates[i])
3. 避免重複選取之前選過的
   candidates[i:]

Time: O(n＾2), Space: O(n)
class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        if target < min(candidates):
                return []
        res = []
        for i in range(len(candidates)):
            if candidates[i]==target:
                res.append([target])
            else:
                for prev in self.combinationSum(candidates[i:],target-candidates[i]):
                    res.append([candidates[i]] + prev)
        return res

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