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：首先很容易用两个指针t1，t2做到前后结点的交换，然后再用一个结点来记录前一个，然后就能处理了

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        if (head == NULL) return NULL;
        if (head->next == NULL) return head;

        ListNode *t1, *t2, *pre;
        t1 = head;
        pre = head;
        while (t1 != NULL && t1->next != NULL) {
            t2 = t1->next;
            t1->next = t2->next;
            t2->next = t1;
            if (pre != head) 
                pre->next = t2;
            else head = t2;
            pre = t1;
            t1 = t1->next;
        }

        return head;
    }
};