39. Combination Sum

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本文深入探讨了LeetCode上的组合总和问题，提供了一种基于记忆化的递归解决方案。通过使用候选数字集和目标数，寻找所有可能的组合，使得组合内数字之和等于目标数。文章详细解释了算法思路，包括如何避免重复组合，以及如何利用记忆化减少重复计算。

https://leetcode.com/problems/combination-sum/

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

我的想法就是coin change的想法，每次金额减去coin币值，在dic里面找amount-coin对应的方法列表，然后每个列表加上当前币值，记得要排序确保唯一性~

class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        mem=collections.defaultdict(list)
        for i in range(target+1):
            for candidate in candidates:
                if candidate>i:
                    continue
                elif candidate==i:
                    mem[i].append([candidate])
                elif candidate<i:
                    tmp=mem[i-candidate]
                    for tmplist in tmp:
                        pair=[candidate]+tmplist
                        if sorted(pair) not in mem[i]:
                            mem[i].append(sorted(pair))
        return mem[target]