40.组合总和II

class Solution:
    def back(self,num,target,start,path,res,used):
        if sum(path) == target:
            res.append(path[:])
            return 
        for i in range(start,len(num)):
            # 因为回溯完，used【i】又变为False，所以当下一个元素跟上一个元素相等时，其实used【i-1】是False的
            if i > start and num[i] == num[i-1] and not used[i-1]:
                continue
            if  sum(path) + num[i] > target:
                break
            path.append(num[i])
            used[i] = True
            self.back(num,target,i+1,path,res,used)
            used[i] = False
            path.pop()
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        candidates.sort()
        res = []
        result = []
        used = [False]*(len(candidates))
        self.back(candidates,target,0,[],res,used)
        return res
        

40. 组合总和 II - 力扣（LeetCode）

一定要掌握去重方法！