【链表】Swap Nodes in Pairs（三指针）

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;
};