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• 842. 排列数字
• • 题意
  • 思路
  • 代码

842. 排列数字

• 题意

• 思路

  DFS爆搜+回溯
  每次遍历这n个数，看哪个数空闲，若空闲就接着DFS，并标记状态

• 代码

  '''
  Author: NEFU AB-IN
  Date: 2022-03-01 23:17:33
  FilePath: \ACM\Acwing\842.py
  LastEditTime: 2022-03-01 23:26:06
  '''
  N = 10
  path, st = [0] * N, [0] * N
  
  
  def dfs(x):
      if x == n + 1:
          for i in range(1, n + 1):
              print(path[i], end=" ")
          print()
          return
      for i in range(1, n + 1):
          if not st[i]:
              path[x] = i
              st[i] = 1
              dfs(x + 1)
              st[i] = 0
  
  
  n = int(input())
  dfs(1)