Leetcode: Reorder List

url :

https://leetcode.com/problems/reorder-list/description/

描述

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}.

代码

class Solution {        public void reorderList(ListNode head) {            if(head == null) return;            ListNode p = head;            int count = 0;            while(p!=null){                p=p.next;                count++;            }            ListNode[] points = new ListNode[count];            p = head;            count = 0;            while(p!=null){                points[count++] = p;                p=p.next;            }            int i = 0, j = count-1;            while(i < j) {                points[i].next = points[j];                points[j--].next = points[++i];            }            points[i].next = null;        }    }

上面的方法总是感觉有点不正宗，下面给出链表的操作，大概就是把链表的后半段作为单独的一段链表，并进行反转，然后在把两个合成一个就行啦，代码如下：

public void reorderList3(ListNode head) {            if(head==null) return;            ListNode fast = head, slow = head;            while(fast!=null && fast.next!=null){                fast = fast.next.next;                if(fast!=null){                    slow = slow.next;                }            }            ListNode pre = null;            ListNode p = slow.next;            ListNode post = p==null ? null : p.next;            while(post!=null){                p.next = pre;                pre = p;                p=post;                post = post.next;            }            if(p!=null)                p.next = pre;            slow.next = null;            ListNode h1 = head;            ListNode h2 = head.next;            ListNode m = p;            while(h1!=null && m!=null) {                h1.next = m;                m = m.next;                h1 = h1.next;                h1.next = h2;                h2 = h2 == null? null : h2.next;                h1 = h1.next;            }        }