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}.

思路: 1.找中点的一个点； 2. reverse后半部分； 3. 连接两个linkedlist

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null) {
            return;
        }
        ListNode middle = findMiddle(head);
        ListNode l1 = head;
        ListNode l2 = middle.next;
        middle.next = null;
        
        l2 = reverseLinkedList(l2);
        
        while(l1 != null && l2 != null) {
            ListNode l1next = l1.next;
            ListNode l2next = l2.next;
            
            l1.next = l2;
            l2.next = l1next;
            
            l1 = l1next;
            l2 = l2next;
        }
    }
    
    public ListNode findMiddle(ListNode node) {
        if(node == null || node.next == null) {
            return node;
        }
        ListNode slow = node;
        ListNode fast = node;
        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    
    public ListNode reverseLinkedList(ListNode head) {
        if(head == null || head.next == null) {
            return head;
        }
        ListNode dummpy = new ListNode(-1);
        dummpy.next = head;
        ListNode cur = dummpy.next;
        while(cur != null && cur.next != null) {
            ListNode curnext = cur.next;
            cur.next = curnext.next;
            curnext.next = dummpy.next;
            dummpy.next = curnext;
        }
        return dummpy.next;
    }
}