Question

这里我的问题是我有一个1-> 2-> 3-> 4-> 5的链表

我的概念是打印它们，如1-> 5-> 2-> 4-> 3

这意味着首先开始然后结束

我该怎么做？

我有一个想法，首先我在空节点中占用一个空节点，我将保持开始节点和

之后我将遍历到最后一个节点将保留在那里我的大脑停止

任何人都可以指导我这样做吗？提前谢谢

def mutate_linked_list(head):
   #node creation
   s = t = Node()
   t.next = head
   head = head.next
   while head is None:
        t.next = head
        head = head.next
        # here i used recursion concept
        mutate_linked_list(head)
  return head

但它没有用......

Answer 1

[head[0]]+[head[-1]] + head[1:-1]

Answer 2

def mutate_linked_list(head):
    # return early for an size 1 list
    if head.next is None:
        return head

    # find the last and penultimate nodes
    penultimate = head
    while penultimate.next.next is not None:
        penultimate = penultimate.next
    last = penultimate.next

    # detach the last node from the end
    penultimate.next = None

    # prepend it to the second node
    last.next = head.next.next

    # and reconnect the head
    head.next = last

    return head

Answer 3

from collections import deque
def mutate_linked_list(head):
   mydeque = deque(head)
   try:
     while True:
       yield mydeque.popleft()
       yield mydeque.pop()
   except IndexError:
     pass

for i in mutate_linked_list(range(1,6)):
  print(i)

你可以显然做一个列表和.append而不是产生价值。

Answer 4

就像练习......

def mix(head):
    # Walk to the end of the list
    last = head
    secondlast = None
    while last.next:
        secondlast = last
        last = last.next

    if secondlast:
         # move last element to second place
         secondlast.next = None
         last.next = head.next
         head.next = last

         # mix the rest of the list (recursion)
         mix(last.next)

Answer 5

怎么样：

from itertools import izip_longest
def mutate_linked_list(head):
    m = len(head)/2 # Get the middle position
    return [item for pair in izip_longest(head[:m+1], head[:m:-1])
                 for item in pair
                 if item is not None]

然后，你准备好了：

head = [1,2,3,4,5]
mutate_linked_list(head)

[1, 5, 2, 4, 3]

它也适用于包含偶数个元素的列表......

head = [1,2,3,4,5,6]
mutate_linked_list(head)

[1, 6, 2, 5, 3, 4]

链表在python中的开头和结尾

5 个答案: