【Leetcode】Reorder List

最新推荐文章于 2025-08-11 15:06:51 发布

转载最新推荐文章于 2025-08-11 15:06:51 发布 · 180 阅读

文章标签：

#leetcode #Java #ListNode

本文介绍了一种在不改变节点值的情况下，对单链表进行重新排序的方法。具体实现包括两个关键步骤：首先通过快慢指针找到链表中点，并将链表分为两部分；然后反转第二部分链表并将其与第一部分交错合并。

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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}.

Java:

http://blog.youkuaiyun.com/linhuanmars/article/details/21503215

public class Solution {
    public void reorderList(ListNode head) {
       if(head==null||head.next==null) return ;
       ListNode walker=head;
       ListNode runner=head;
       while(runner!=null&&runner.next!=null)
       {
           walker=walker.next;
           runner=runner.next.next;
       }
       ListNode head1=head;
       ListNode head2=walker.next;
       walker.next=null;
       head2=reverse(head2);//iteration reverse
       //merge below
       while(head1!=null&&head2!=null)
       {
           ListNode next=head2.next;
           head2.next=head1.next;
           head1.next=head2;
           head1=head2.next;
           head2=next;
       }
    }
    
    // iteration reverse
    private ListNode reverse(ListNode head)
    {
        ListNode pre=null;
        ListNode cur=head;
        while(cur!=null)
        {
            ListNode next=cur.next;
            cur.next=pre;
            pre=cur;
            cur=next;
        }
        return pre;
    }

}

Recursive Reverse List:

 //recursive reverse
    public ListNode rReverse(ListNode head)
    {
        if(head==null||head.next==null) return head;
        return rReverse(head,head.next);
    }
    private ListNode rReverse(ListNode current,ListNode next)
    {
        if(next==null) return current;
        ListNode newHead=rReverse(current.next, next.next);
        next.next=current;
        current.next=null;
        return newHead;
    }