40. Combination Sum II

这一题的思路是，首先需要排序，因为数字不可以重复，所以只能往后面找数字，同时因为candidates本身是有可能有重复的，所以在同一个位置上，需要判断一下是不是重复，重复就没有必要再来加一次了。

class Solution(object):
    def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        candidates.sort()
        res = []
        self.surpport(0, candidates, target, [], 0, res)
        return res
    
    def surpport(self, start, candidates, target, temp, count, res):
        if count == target:
            res.append(temp)
            return
        elif count > target:
            return
        else:
            new = []
            for i in range(start, len(candidates)):
                if i > 0 and candidates[i-1] in new and candidates[i] == candidates[i-1]:
                    continue
                new.append(candidates[i])
                self.surpport(i+1, candidates, target, temp + [candidates[i]], count+candidates[i], res)