题目描述：
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes' values.
For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.
本题算是链表中很有特点的一道题目。对于链表1->2->3->4->5->6 ，要变成1->6->2->5->3->4，
即第i个节点指向倒数第i个节点，而倒数第i个节点，指向第i+1个节点。
解法一：
使用前后两个指针p和q。
具体实现步骤在注释中有详细说明，在此不再赘述。
由于时间复杂度为O(N^2)，不够高效导致会超时。无法通过OJ的测试数据。
实现代码：
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public void ReorderList(ListNode head)
{
if(head == null || head.next == null || head.next.next == null){
return;
}
var p = head;
var q = head;
while(q.next.next != null){
while(p.next.next != null){
p = p.next;
}
// point head to last
var t = q.next;
q.next = p.next;
// point last to 2nd and set the second last to null
p.next.next = t;
// point 2nd last to null
p.next = null;
// reset p and q
p = t;
q = t;
if(q.next == null){
break;
}
}
}
}
解法二：
本实现参考了连接：
http://www.acmerblog.com/reorder-list-leetcode-6088.html
1.使用slow和fast指针将链表分为两部分，part1和part2 ，假设链表为1->2->3->4->5->6->7->8, part1 = {1->2->3->4} , part2= {5->6->7->8}
2.然后对part2逆置,即8->7->6->5
3.然后分别将part1[0]->part2[0], part2[0]->part1[1], part1[1]->part2[1]...
即，对于i < len - 1
part1[i] -> part2[i]
part2[i] -> part1[i+1]
i++
实现代码：
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public void ReorderList(ListNode head)
{
if(head == null || head.next == null || head.next.next == null){
return;
}
var slow = head;
var fast = head;
while(fast.next != null && fast.next.next != null)
{
slow = slow.next;
fast = fast.next.next;
}
var mid = slow.next;
var last = mid;
ListNode pre = null;
while(last != null){
ListNode next = last.next;
last.next = pre;
pre = last;
last = next;
}
slow.next = null;
while(head != null && pre != null)
{
var next1 = head.next;
head.next = pre;
pre = pre.next;
head.next.next = next1;
head = next1;
}
}
}