Python version-LeetCode learning: 46. Full arrangement

Given a  sequence  without repeated numbers, return all possible permutations.

Example: Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1 ,2],
  [3,2,1]
]

Source: LeetCode
Link: https://leetcode-cn.com/problems/permutations

Method 1: This method mainly uses the backtracking method,

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        track=[]
        res=[]

        def arrange(nums,track):
            if len(track)==len(nums):
                # 由于list是可变对象，因此，这里要用[:]取到具体的值，再append进res中。
                res.append(track[:])
                return 
            for i in nums:
                if i in track:
                    # 如果nums数值已在搜寻表中，继续下一轮
                    continue
                # 添加路径
                track.append(i)
                arrange(nums,track)
                # 对路径进行回溯，将原添加进的值移除
                track.pop()
        
        arrange(nums,track)
        return res

 Method 2:

It mainly uses the recursive iterative process to perform different splicing of the elements on the list to traverse various possible values.

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        res = []
        def cur(nums, temp):
            if not nums:
                # 将中间结果加入结果集
                res.append(temp)
                return
            for i in range(len(nums)):
                # 利用递归的迭代进行结果遍历
                cur(nums[:i] + nums[i+1:], temp + [nums[i]])
        cur(nums, [])
        return res

 This is the least amount of code I have seen to solve the whole arrangement.

ps: when you have time to make up