题目：
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.
思路：
给链表添加一个临时的头结点, 这样操作更方便。其实大部分链表问题,添加一个头结点，都会简化后面的操作
/** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} head * @return {ListNode} */ var swapPairs = function(head) { var tempHead=new ListNode(0); tempHead.next=head; var pre=tempHead,p=head; while(p&&p.next){ pre.next=p.next; p.next=p.next.next; pre.next.next=p; pre=p; p=p.next; } return tempHead.next; };
原文：http://www.cnblogs.com/shytong/p/5142494.html