Reorder List

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-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}.

分析：先用快慢指针找到链表的中点，然后翻转链表后半部分，再和前半部分组合。需要注意的是把链表分成两半时，前半段的尾节点要置为NULL，翻转链表时也要把尾节点置为NULL。

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(head == null || head.next == null) return;
        ListNode fast = head, slow = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        fast = slow.next;
        slow.next = null;
        fast = reverse(fast);
        while(head != null && fast != null){//merge
            ListNode tmp1 = head.next;
            ListNode tmp2 = fast.next;
            head.next = fast;
            fast.next = tmp1;
            head = tmp1;
            fast = tmp2;
        }
    }
    ListNode reverse(ListNode root){
        ListNode first = null, second = root;
        while(second != null){
            ListNode tmp = second.next;
            second.next = first;
            first = second;
            second = tmp;
        }
        return first;
    }
}