Leetcode: reorder-list

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：

这道题可以分三个步骤来解决。一、找到原链表的中间位置，以这个中间位置为基准将链表分为两段；二、将处在后半段的链表进行反转；三、将两个链表进行融合。

具体的代码如下：

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public static void reorderList(ListNode head) {        if (head == null || head.next == null)            return ;                ListNode mid = findMiddle(head);                ListNode aft = mid.next;        mid.next = null;        ListNode pre = head;                aft = reverse(aft);        merge(pre, aft);    }        public static ListNode reverse(ListNode aft){        ListNode l1 = aft;        ListNode l2 = aft.next;                while(l2 != null){            ListNode tmp = l2.next;            l2.next = l1;            l1 = l2;            l2 = tmp;        }        aft.next = null;                return l1;    }        public static void merge(ListNode pre, ListNode aft){        while (aft != null){            ListNode aftN = aft.next;            ListNode preN = pre.next;                        pre.next = aft;            aft.next = preN;                        aft = aftN;            pre = preN;        }    }        public static ListNode findMiddle(ListNode head){        ListNode slow = head;        ListNode fast = head;                while(fast.next != null && fast.next.next != null){            slow = slow.next;            fast = fast.next.next;        }                return slow;    }}