基于Python实现扑克牌面试题

def cl(n):
  flst = [] #用数组flst定义最后桌上的牌堆顺序 
  for i in range(1, n+1): 
    flst.append([i, i])
  if n%2 == 0: #n为偶数情况
    for t in range(1, int(n/2)+1): #t为循环次数，代表一次取牌放牌顶+牌底的过程
      flst[n-t][0] = 2*t - 1
      flst[t-1][0] = 2*t
  else: #n为奇数情况
    flst[0][0]= 1 #新牌堆首张牌必定为原牌堆的最后一张
    for t in range(1, n//2+1):
      flst[n-t][0] = 2*t
      flst[t][0] = 2*t + 1
  olst = sorted(flst, key=lambda x: x[0], reverse=True) #对二维数组进行排序
  res = [i[1] for i in olst] #求得原牌组牌号
  return res

 print(cl(10))
 print(cl(11))

[5, 6, 4, 7, 3, 8, 2, 9, 1, 10]
[6, 7, 5, 8, 4, 9, 3, 10, 2, 11, 1]