class Solution: def permute(self,nums): result = [] # 回溯函数,path 是当前构建的排列,used 是标记已使用的数字 def backtrack(path, used): if len(path) == len(nums): result.append(path[:]) # 找到一个完整的排列 return for i in range(len(nums)): if not used[i]: # 做选择 path.append(nums[i]) used[i] = True # 递归构造剩下的排列 backtrack(path, use...