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Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4
, you should return the list as 2->1->4->3
.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
需要运用fakehead来指向原指针头,防止丢链,用两个指针,pre始终指向需要交换的pair的前面一个node,cur始终指向需要交换的pair的第一个node。
然后就进行正常的链表交换,和指针挪动就好。
当链表长度为奇数时,cur.next可能为null;
当链表长度为偶数时, cur可能为null。
所以把这两个情况作为终止条件,在while判断就好,最后返回fakehead.next。
链表交换死活看不懂,需要人帮我讲下。
Java code:
public class Solution { public ListNode swapPairs(ListNode head) { if(head == null || head.next == null){ return head; } ListNode fakehead = new ListNode(0); fakehead.next = head; ListNode pre = fakehead; ListNode cur = head; while(cur!= null && cur.next != null) { pre.next = cur.next; cur.next = cur.next.next; pre.next.next = cur; pre = cur; cur = cur.next; } return fakehead.next; } }
Reference:
1. http://bangbingsyb.blogspot.com/2014/11/leetcode-swap-nodes-in-pairs.html
2. http://www.cnblogs.com/springfor/p/3862030.html
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原文地址:http://www.cnblogs.com/anne-vista/p/4963680.html