Leetcode Reorder List 链表重排序

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}.

分析：

该题目实际上是几个链表问题的综合，用到的方法有找链表中点，反转链表和合并链表，只需要把这几个算法稍微修改适合本题的情景即可。

1. 把链表分成两部分，当链表长度为偶数时，平均分，当链表长度为奇数时， L0链表多一个元素。用找链表中点的方法实现。

2. 对Ln链表进行反转。

3. 把两个链表进行合并。

Java代码实现：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public void reorderList(ListNode head) {        if(head==null || head.next==null || head.next.next==null)            return;                ListNode mid = findmid(head);        ListNode start = reverse(mid.next);        mid.next = null;                merge(head, start);    }    private ListNode findmid(ListNode head)    {        ListNode fast = head.next;        ListNode slow = head;        while(fast!=null && fast.next!=null)        {            fast = fast.next.next;            slow = slow.next;        }        return slow;    }    private ListNode reverse(ListNode head)    {        ListNode node = head;        ListNode pre = node.next;        node.next = null;        while(pre!=null)        {            ListNode temp = pre.next;            pre.next = node;            node = pre;            pre = temp;        }        return node;    }    private void merge(ListNode l1, ListNode l2)    {        ListNode dummy = new ListNode(0);        int i = 0;        while(l2!=null)        {            if(i%2==0)            {                dummy.next = l1;                l1 = l1.next;            }            else            {                dummy.next = l2;                l2 = l2.next;            }            dummy = dummy.next;            i++;        }        if(l1!=null)            dummy.next = l1;    }}