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.
题目要求成对翻转指针，难度一般，要求使用常数的空间，应当使用迭代的方法而非递归的方法，这里我们加上一个头结点dummy，这样一来即使翻转的节点涉及到头结点，需要修改头指针，我们也可以做到和一般节点一样处理，我们设置了一个指针pre指向“一对”节点，一个指针t指向一对节点的第二个节点
【图解稍后补充】
代码如下：
1 /**
2 * Definition for singly-linked list.
3 * struct ListNode {
4 * int val;
5 * ListNode *next;
6 * ListNode(int x) : val(x), next(NULL) {}
7 * };
8 */
9 class Solution {
10 public:
11 ListNode* swapPairs(ListNode* head) {
12 ListNode *dummy = new ListNode(-1), *pre = dummy;
13 dummy->next = head;
14 while (pre->next && pre->next->next)
15 {
16 ListNode *t = pre->next->next;
17 pre->next->next = t->next;
18 t->next = pre->next;
19 pre->next = t;
20 pre = t->next;
21 }
22 return dummy->next;
23 }
24 };
