【Leetcode】【python】Reorder List 重排链表

题目大意

将单向链表L0→L1→…→Ln-1→Ln转化为L0→Ln→L1→Ln-1→L2→Ln-2→…的形式,也就是从头部取一个节点,从尾部取一个节点,直到将原链表转化成新的链表。

解题思路

  1. 去中间节点,将链表分为两段.
  1. 翻转后一段
  2. 拼接

    代码

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class Solution:
def reorderList(self, head):
"""
:type head: ListNode
:rtype: void Do not return anything, modify head in-place instead.
"""
if not head:
return
# split {1,2,3,4,5} to {1,2,3}{4,5}
fast = slow = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
head1 = head
head2 = slow.next
slow.next = None
# reverse the second {4,5} to {5,4}
cur, pre = head2, None
while cur:
nex = cur.next
cur.next = pre
pre = cur
cur = nex
# merge
cur1, cur2 = head1, pre
while cur2:
nex1, nex2 = cur1.next, cur2.next
cur1.next = cur2
cur2.next = nex1
cur1, cur2 = nex1, nex2

总结