143. 重排链表

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

方法1：转化为线性表，再按照要求链接节点

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/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reorderList(head *ListNode) {
	var nums []*ListNode
	dummy := head
	for dummy != nil {
		nums = append(nums, dummy)
		dummy = dummy.Next
	}
	  
	l, r := 0, len(nums)-1
	for l<r {
		nums[l].Next = nums[r]
		l++
		if l == r {
			break
		}
		nums[r].Next = nums[l]
		r--
	}
	nums[l].Next = nil
	return
}

方法2：从链表中间切断，后半段逆序，然后再把二者合并
链表操作秀肌肉

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/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reorderList(head *ListNode) {
	if head == nil {
		return
	}
	mid := findMiddle(head)
	l1 := head
	l2 := mid.Next
	mid.Next = nil
	nl2 := reverse(l2)
	merge(l1, nl2)
}
  
func findMiddle(head *ListNode) *ListNode {
	slow, fast := head, head
	for fast.Next != nil && fast.Next.Next != nil {
		slow = slow.Next
		fast = fast.Next.Next
	}
	return slow
}
  
func reverse(head *ListNode) *ListNode {
	var pre, cur *ListNode = nil, head
	for cur != nil {
		nxt := cur.Next
		cur.Next = pre
		pre = cur
		cur = nxt
	}
	return pre
}
  
func merge(h1, h2 *ListNode) *ListNode {
	dummy := &ListNode{}
	t := dummy
	for h1 != nil && h2 != nil {
		t.Next = h1
		t = t.Next
		h1 = h1.Next
		  
		t.Next = h2
		t = t.Next
		h2 = h2.Next
	}
	if h1 != nil {
		t.Next = h1
	} else if h2 != nil {
		t.Next = h2
	}
	return dummy.Next
}