交换单链列表的最后两个节点

本文介绍了交换单链列表的最后两个节点的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

如何交换链表的最后两个节点?我正在尝试使用一个辅助节点，因为我认为有必要避免在此过程中丢失"一个节点...

How can I swap the last two nodes of a linked list? I'm trying to use a helper node as I think it's needed to avoid 'losing' a node in the process...

... Node node3 = new Node("Hi", null) ; Node node4 = new Node("Hello", null) ; ... // swap node3 & node4 Node temp = node3.succ ; node3.succ = null ; // this should be the last node now, so i set its pointer to null node2.succ = temp ; // the second's node successor becomes what used to be the last node temp = node4 ; // not sure how to use temp here. what should it point to if at anything?

我认为我做错了，有什么提示吗?

I think I'm doing this wrong, any hints?

推荐答案

假设您有一个链表 A->B->C ，并且您想交换 B 和 C :

Suppose you have a linked list A -> B -> C, and you want to swap B and C:

设置T * = B(将B存储在某个地方)

设置A.next = C

设置T * .next = C.next(这可以从仅在列表末尾进行操作而得到概括)

设置C.next = T *