Reorder List

题目地址：
https://leetcode.com/problems/reorder-list/description/

题目：

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}.

解题思路：
这道题主要综合了三个知识点：链表找中点，反转链表和merge链表。

代码：

public class ReorderList {

    public static void main(String[] args){
        ReorderList reorderList = new ReorderList();
        ListNode n1 = new ListNode(1);
        ListNode n2 = new ListNode(2);
        ListNode n3 = new ListNode(3);
        ListNode n4 = new ListNode(4);
        ListNode n5 = new ListNode(5);
        ListNode n6 = new ListNode(6);
        ListNode n7 = new ListNode(7);
        n1.next = n2;
        n2.next = n3;
        n3.next = n4;
        n4.next = n5;
        n5.next = n6;
        n6.next = n7;
        reorderList.reorderList(n1);
        while(n1 != null){
            System.out.print(n1.val + "-");
            n1 = n1.next;
        }
    }

    public void reorderList(ListNode head) {
        if(head == null || head.next == null || head.next.next == null){
            return;
        }
        ListNode fast = head;
        ListNode slow = head;
        // this slow pointer will stay at the mid point for list with odd number        // the slow pointer will stay at the node closer to the head among the mid two node        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode head1 = head;
        ListNode oldHead2 = slow.next;
        slow.next = null;
        ListNode head2 = reverse(oldHead2);
        merge(head1, head2);
    }

    private void merge(ListNode head1, ListNode head2) {
        ListNode dummy = new ListNode(-1);
        ListNode curr = dummy;
        int count = 0;
        while(head1 != null && head2 != null){
            if(count % 2 == 0){
                curr.next = head1;
                head1 = head1.next;
            }
            else{
                curr.next = head2;
                head2 = head2.next;
            }
            count++;
            curr = curr.next;
        }
        if(head1 != null){
            curr.next = head1;
        }
        if(head2 != null){
            curr.next = head2;
        }
        head1 = dummy.next;
    }

    // iterative way    public ListNode reverse(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode curr = head;
        ListNode next = curr.next;
        curr.next = null;
        while(next != null){
            ListNode nextnext = next.next;
            next.next = curr;
            curr = next;
            next = nextnext;
        }
        return curr;
    }


    // recursion might overflow//    public ListNode reverse(ListNode head) {//        if(head == null || head.next == null){//            return head;//        }//        ListNode next = head.next;//        ListNode newHead = reverse(next);//        next.next = head;//        head.next = null;//        return newHead;//    }
}