combination sum ii

Given a collection of candidate numbers (C) and a target number (T),
find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Example 
Given candidate set [10,1,6,7,2,1,5] and target 8,

A solution set is:
[
  [1,7],
  [1,2,5],
  [2,6],
  [1,1,6]
]

这个题最难的地方在于如何处理重复数字
如果不去重，就会出现1(1), 2, 5, 1(2), 2, 5 的情况

解决办法就是

只有看到1(1) 在结果中时才使用1(2)
这样这个1(2), 2, 5就不会出现，而1, 1, 6会出现

⚠️如果直接在看到重复数字就跳过，那么1,1,6就不会出现了

code

class Solution:
    """
    @param candidates: Given the candidate numbers
    @param target: Given the target number
    @return: All the combinations that sum to target
    """

    def combinationSum2(self, candidates, target):
  
        candidates.sort()  # sort it first !
        self.result = []
        used = [False] * len(candidates)
        self.dfsHelper(candidates, target, [], 0, used)

        return self.result

    # @param: candidates: list, target: int, path: list, index: int
    # @return: None
    def dfsHelper(self, candidates, target, path, index, used):
        # base
        if target == 0:
            self.result.append(list(path))
            return
        if target < 0:
            return

        # divide
        for i in range(index, len(candidates)):
            # only use the following same number when previous same number is used
            if i != 0 and candidates[i] == candidates[i - 1]:
                if not used[i - 1]:
                    continue

            path.append(candidates[i])
            used[i] = True
            self.dfsHelper(candidates, target - candidates[i], path, i + 1, used)
            used[i] = False
            path.pop()