LeetCode：reorder-list链表重排序_将给定的单链表: 重新排序为:

题目描述

将给定的单链表L： L 0→L 1→…→L n-1→L n,

重新排序为： L 0→L n →L 1→L n-1→L 2→L n-2→…

要求使用原地算法，并且不改变节点的值

例如：

对于给定的单链表{1,2,3,4}，将其重新排序为{1,4,2,3}.

Given a singly linked list L: L 0→L 1→…→L n-1→L n,

reorder it to: L 0→L n →L 1→L n-1→L 2→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}.

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解题思路：

（1）利用快慢指针找到链表中间结点

（2）以中间节点为准将链表拆分成两半，反转中间结点后的后半部分链表

（3）合并前半部分链表和反转后的后半部分链表

Java代码实现

/**
 * 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) {
        if(head == null || head.next == null)
            return;
        //快慢指针寻找中间结点
        ListNode slow = head, fast = head;
        while(fast.next != null && fast.next.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        ListNode after = slow.next;
        //断链，拆分链表
        slow.next = null;
        ListNode pre = null;
        //反转链表
        while(after != null){
            ListNode tmp = after.next;
            after.next = pre;
            pre = after;
            after = tmp;
        }
        //合并链表
        ListNode first = head;
        after = pre;
        while(first != null && after != null){
            ListNode first_tmp = first.next;
            ListNode after_tmp = after.next;
            first.next = after;
            first = first_tmp;
            after.next = first;
            after = after_tmp;  
        }
    }
}