1 题目
2 思想
3 代码
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reorderList(self, head: ListNode) -> None:
"""
Do not return anything, modify head in-place instead.
"""
mid = self.splitList(head)
# 将 mid 开头的这部分链表翻转
head2 = self.reverseList(mid)
return self.mergeList(head,head2)
# 二分链表
def splitList(self,head):
slow = fast = head
while(fast.next and fast.next.next):
fast = fast.next.next
slow = slow.next
tmp = slow.next
slow.next = None
return tmp
# 翻转链表
def reverseList(self,head):
pre = None
start = head
while(start):
tmp = start.next
start.next = pre
pre = start
start = tmp
return pre
# 交叉合并
def mergeList(self,head1,head2):
cnt = 0
tail = head = head1 # 最后返回的头节点
head1 = head1.next
while(head1 and head2):
if cnt % 2 == 0:
tail.next = head2
head2 = head2.next
else:
tail.next = head1
head1 = head1.next
tail = tail.next
cnt += 1
if head1:
tail.next = head1
if head2:
tail.next = head2
return head
