2021-08-07
给定一个单链表 L 的头节点 head ,单链表 L 表示为:
L0 → L1 → … → Ln-1 → Ln
请将其重新排列后变为:
L0 → Ln → L1 → Ln-1 → L2 → Ln-2 → …
不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
示例 1:
输入: head = [1,2,3,4]
输出: [1,4,2,3]
示例 2:
输入: head = [1,2,3,4,5]
输出: [1,5,2,4,3]
提示:
链表的长度范围为 [1, 5 * 104]
1 <= node.val <= 1000
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reorder-list
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解题思路
# 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.
"""
fast = slow = a = head
while fast and fast.next:
fast = fast.next.next
slow = slow.next
b = slow.next if fast else slow
pre = None
cur = b
while cur:
hl = cur.next
cur.next = pre
pre = cur
cur = hl
b = pre
while a:
a.next, b = b, a.next
a = a.next
return head