剑指 Offer II 026. 重排链表

题目链接

力扣

题目描述

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

解题思路

题目并不难，代码量比较大，首先将链表从中间断开，然后把后半段反转，然后再合并即可

代码

Python

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        prev = None
        cur = head
        while cur:
            next = cur.next
            cur.next = prev
            prev = cur
            cur = next
        return prev

    def findMidNode(self, head: ListNode) -> ListNode:
        slow, fast = head, head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
        mid = slow.next
        slow.next = None
        return mid

    def mergeList(self, head1: ListNode, head2: ListNode) -> ListNode:
        l1, l2 = head1, head2
        while l1 and l2:
            next1 = l1.next
            next2 = l2.next
            l1.next = l2
            l1 = next1
            l2.next = l1
            l2 = next2
        return head1

    def reorderList(self, head: ListNode) -> None:
        mid = self.findMidNode(head)
        tail = self.reverseList(mid)
        self.head = self.mergeList(head, tail)

Go

func reorderList(head *ListNode) {
	if head.Next == nil {
		return
	}
	midNode := findMidNode(head)
	tail := reverseListNode(midNode)
	l1, l2 := head, tail
	next1, next2 := l1.Next, l2.Next
	for l2 != nil {
		l1.Next = l2
		l1 = next1
		if next1 != nil {
			next1 = next1.Next
		}
		l2.Next = l1
		l2 = next2
		if next2 != nil {
			next2 = next2.Next
		}
	}
}

func reverseListNode(head *ListNode) *ListNode {
	var prev *ListNode
	cur := head
	for cur != nil {
		next := cur.Next
		cur.Next = prev
		prev = cur
		cur = next
	}
	return prev
}

func findMidNode(head *ListNode) *ListNode {
	slow, fast := head, head
	for fast.Next != nil && fast.Next.Next != nil {
		slow = slow.Next
		fast = fast.Next.Next
	}
	mid := slow.Next
	slow.Next = nil
	return mid
}