【leetcode】Reorder List

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}.

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题解：看起来最后变换得到的链表顺序很奇怪，起始很简单，就是把链表从中间折断，后半部分反转后和前半部分做归并即可。不过这个归并不是归并排序每次从两个链表中取较小的值，而是两个链表相互交错即可。

例如1->2->3->4，从中间折断得到两个链表1->2和3->4，后面一个链表反转得到4->3，然后和第一个链表交错归并得到1->3->2->4;

再例如1->2->3->4->5，从中间折断得到两个链表1->2->3和4->5，后一个链表反转得到5->4，和第一个链表交错归并得到1->4->2->5->3;

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 {
     private ListNode reverse(ListNode head){
         ListNode newHead = null;
         while(head != null){
             ListNode temp = head.next;
             head.next = newHead;
             newHead = head;
             head = temp;
         }
         return newHead;
     }
     
     private void newMerge(ListNode head1,ListNode head2){
         ListNode newHead = new ListNode(0);
         
         while(head1 != null && head2 != null){
             newHead.next = head1;
             head1 = head1.next;
             newHead = newHead.next;
             newHead.next = head2;
             head2 = head2.next;
             newHead = newHead.next;
         }
         if(head1 != null)
             newHead.next = head1;
         if(head2 != null)
             newHead.next = head2;
     }
     private ListNode findMiddle(ListNode head){
          ListNode slow = head;
          ListNode fast = head.next;
          while(fast != null && fast.next != null)
          {
              slow = slow.next;
              fast = fast.next.next;
          }
          return slow;
      }
    public void reorderList(ListNode head) {
        if(head == null || head.next == null)
            return;
        
        ListNode mid = findMiddle(head);
        ListNode tail = reverse(mid.next);
        mid.next = null;
        
        newMerge(head, tail);
    }
}