【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;

};

巴特西

【链表】Swap Nodes in Pairs（三指针）

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