给定一个单链表 L：L0→L1→…→Ln-1→Ln ， 将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…
你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:
给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例 2:
给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
解：纯手打，竟然一次ac,讶于自己编码能力提升！先将链表切分两半，后半反转，然后合并。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void reorderList(ListNode head) {
if (head == null) {
return;
}
ListNode realRtn=head;
ListNode fast = head;
ListNode slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
ListNode halfTail = null;
if (fast != null) {//链表长度为奇数
halfTail = slow.next;
} else {//链表长度为偶数
halfTail = slow;
}
ListNode revList = rev(halfTail);
while (revList != null) {
ListNode tmp1 = revList.next;
ListNode tmp2 = realRtn.next;
revList.next = realRtn.next;
realRtn.next = revList;
revList = tmp1;
realRtn = tmp2;
}
realRtn.next=null;
}
private ListNode rev(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode newList = null;
while (head != null) {
ListNode tmp = head.next;
head.next = newList;
newList = head;
head = tmp;
}
return newList;
}
}
