Swap Nodes in Pairs 



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：题目比較简单，会链表反转的都能够做，思路也差点儿相同。不多说。上代码。

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) { val = x; }

 * }

 */

public class Solution {

	    public ListNode swapPairs(ListNode head) {

	        ListNode firstHead = new ListNode(0);

	        firstHead.next = head;

	        ListNode pre = firstHead;//定义一个头结点，这样全部的操作都同样

	        ListNode next = null;

	        while(head != null && head.next != null){

	        	//A-B-C-D交换BC,pre=A;B=head;C=next

	            next = head.next;//保存交换的变量C

	            head.next = next.next;//将B指向B的指针指向D

	            pre.next = next;//将A指向B的指针指向C

	            next.next = head;//将C指向D的指针指向B,完毕交换，顺序变为A-C-B-D

	            //为下一循环准备变量

	            pre = head;//将pre变为B

	            head = head.next;//将head指向D

	        }

	        return firstHead.next;

	    }

}