[backtracking] full arrangement II leetcode47

Given a sequence that can contain repeating numbers nums , In any order Returns all non repeating permutations . Example 1： Input ：nums = [1,1,2] Output ：
[[1,1,2], [1,2,1], [2,1,1]] Example 2： Input ：nums = [1,2,3]
Output ：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
link ：https://leetcode.cn/problems/permutations-ii

Ideas ： and lc46 What's different is , Contains duplicate numbers , here , Learn from array sum II The idea of , First, let's talk about sequence sorting , Then every time you traverse for Loop to determine whether the next number is the same as the previous one , Same auto ignore （ prune ） fall , So as to achieve the goal of weight removal

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        self.ret = []

        '''NOTE 1. Thinking about recursive trees （ What are the root and intermediate nodes ） 2. Think about each state of the recursive tree （for The loop condition ） 3. Think about the variables of the recursive tree （ Formal parameters of the backtrace function ） 4. Think about what the leaf nodes of a recursive tree are （ Where to end return And save what ） '''
        def helper(nums:list,his:list):
            if len(nums)==0:
                self.ret.append(his)
                return
            for idx in range(len(nums)):
                if idx!=0 and nums[idx]==nums[idx-1]:   #  Weight removal tips 
                    continue
                helper(nums[:idx]+nums[idx+1:],his+[nums[idx]])
        nums.sort()
        helper(nums,[])
        return self.ret