今天回顾-回溯算法-组合40

注意点&感悟：

• 还是得复习！！多巩固巩固，我可以的！！！！！

题目链接：40. 组合总和 II

自己独立写的代码：

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        res = []
        candidates.sort()
        used = [0] * len(candidates)
        self.backtracking(candidates,target,0,[],0,res,used)
        return res
    
    def backtracking(self,candidates,target,start_index,path,total,res,used):
        # 特殊 终止条件
        if total == target:
            res.append(path[:])
            return
        # 核心
        for i in range(start_index,len(candidates)):
            # 1.去重
            if i>0 and candidates[i] == candidates[i-1] and used[i-1] ==0:  # used未使用
                continue
            total += candidates[i]
            # 2.过滤
            if total > target:
                break
            path.append(candidates[i])
            used[i] = 1
            self.backtracking(candidates,target,i+1,path,total,res,used)
            used[i] = 0
            path.pop()
            total -= candidates[i]

通过截图：