24. Swap Nodes in Pairs（6.30%）

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.
解释：给你一个链表，从表头开始，每两个节点交换顺序，比如1->2->3->4，那么1和2交换顺序，3和4交换顺序，然后把交换顺序之后的这个表的表头返回。要求您的算法应该使用不变的内存（就是说内存不能不断地递增），而且不能改变节点的值，它们只是顺序改变了。

/**
 * 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) {
        if (head == null || head.next == null) {
            return head;
        } 

        ListNode firstNode = head;
        ListNode secondNode = head.next;
        ListNode thirdNode = head.next.next;

        firstNode.next = swapPairs(thirdNode);
        secondNode.next = firstNode;

        return secondNode;
    }
}