问题：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.

这个问题其实不难理解，就是要在链表中不停做元素位置的交换。我们需要一个临时指针和两个指针分别之前当前节点和下一节点。交换两者的操作比较基础，把前一节点的next设置为后节点的next，然后后节点的next设置成前节点。昨晚这一步后，整体后移。直到链表尽头。代码如下：

/**
 * 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) {
    ListNode **pp = &head, *a, *b;
    while ((a = *pp) && (b = a->next)) {
        a->next = b->next;
        b->next = a;
        *pp = b;
        pp = &(a->next);
    }
    return head;
    }
};