[Leetcode] reorder list

Given a singly linked listL:L0 →L1 →... →LN-1 →LN,
Reorder it: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{1,4,2,3}.

 

Question: It seems that the sequence of the chain table obtained after the last transformation is very strange. The starting point is very simple. It is to break the chain table from the center, reverse the half, and merge with the first half. However, this sort is not to take a small value from the two linked lists each time, but to staggered the two linked lists.

For example, 1-> 2-> 3-> 4, break the link in the middle to get two linked lists 1-> 2 and 3-> 4, and the next linked list is reversed to 4-> 3, then it is staggered with the first linked list to get 1-> 3-> 2-> 4;

For example, 1-> 2-> 3-> 4-> 5, break two linked lists from the center, 1-> 2-> 3 and 4-> 5, the next linked list is reversed to 5-> 4, and the first linked list is staggered and merged to 1-> 4-> 2-> 5-> 3;

The Code is as follows:

 1 /** 2  * Definition for singly-linked list. 3  * class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { 7  *         val = x; 8  *         next = null; 9  *     }10  * }11  */12 public class Solution {13      private ListNode reverse(ListNode head){14          ListNode newHead = null;15          while(head != null){16              ListNode temp = head.next;17              head.next = newHead;18              newHead = head;19              head = temp;20          }21          return newHead;22      }23      24      private void newMerge(ListNode head1,ListNode head2){25          ListNode newHead = new ListNode(0);26          27          while(head1 != null && head2 != null){28              newHead.next = head1;29              head1 = head1.next;30              newHead = newHead.next;31              newHead.next = head2;32              head2 = head2.next;33              newHead = newHead.next;34          }35          if(head1 != null)36              newHead.next = head1;37          if(head2 != null)38              newHead.next = head2;39      }40      private ListNode findMiddle(ListNode head){41           ListNode slow = head;42           ListNode fast = head.next;43           while(fast != null && fast.next != null)44           {45               slow = slow.next;46               fast = fast.next.next;47           }48           return slow;49       }50     public void reorderList(ListNode head) {51         if(head == null || head.next == null)52             return;53         54         ListNode mid = findMiddle(head);55         ListNode tail = reverse(mid.next);56         mid.next = null;57         58         newMerge(head, tail);59     }60 }