LeetCode 46. 全排列

给定一个 没有重复 数字的序列，返回其所有可能的全排列。

示例:

输入: [1,2,3]
输出:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        if len(nums) <= 1:  # 递归终止条件
            return [nums]
        res = []
        for idx, num in enumerate(nums):
            res_nums = nums[:idx] + nums[idx + 1:]  # 确定剩余元素
            for j in self.permute(res_nums):  
                res.append([num] + j)
        return res

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        ans = []
        cur = []
        visit = set()
        self.permute_solver(nums, ans, cur, visit)
        return ans 

    def permute_solver(self,nums,ans,cur,visit):
        if len(cur)==len(nums):
            ans.append(cur[:])
            return
        for i in range(len(nums)):
            if nums[i] in visit:
                continue
            cur.append(nums[i])
            visit.add(nums[i])
            self.permute_solver(nums,ans,cur,visit)
            num = cur.pop()
            visit.remove(num)
        return 

    def permute(self, nums):
        def backtrack(first = 0):
            print(nums)
            # 所有数都填完了
            if first == n:  
                res.append(nums[:])
            for i in range(first, n):
                # 动态维护数组
                nums[first], nums[i] = nums[i], nums[first]
                # 继续递归填下一个数
                backtrack(first + 1)
                # 撤销操作
                nums[first], nums[i] = nums[i], nums[first]
        
        n = len(nums)
        res = []
        backtrack()
        return res