Reorder List

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}.

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if(head == null){
            return;
        }
        if(head.next == null || head.next.next == null){//有两个以上的节点翻转才有意义
            return;
        }
        ListNode p1 = head;
        ListNode p2 = head;
        while(p2.next != null && p2.next.next != null){
            p1 = p1.next;
            p2 = p2.next.next;
        }
        //p1的位置就是中间位置
        ListNode second = new ListNode(-1);
        ListNode p = p1.next;
        p1.next = null;
        while(p != null){
            ListNode q = p.next;
            p.next = second.next;
            second.next = p;
            p = q;
        }
        p1 = head;
        //奇数时前半条链多一个元素，偶数时前后链元素个数相同
        second = second.next;//second有一个头元素，这个节点不存储任何信息，只是来标志是头结点,这样操作方便
        while(second != null){
            p2 = p1.next;
            p1.next = second;
            second = second.next;
            p1.next.next = p2;
            p1 = p2;
        }
    }
}

Runtime:  500 ms