Swap Nodes in Pairs -- leetcode

题目描述：
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.

大意是：成对的交换链表的结点，1和2交换；3和4交换，5和6交换等。

/**
 * 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 || head->next == NULL) return head;
        ListNode *newList = new ListNode(0);
        newList->next = head;
        head = newList;
       while(head->next != NULL && head->next->next != NULL){
           ListNode *p1 = head -> next;//取出第一个节点结点 
           ListNode *p2 = head -> next -> next;//取出第二个节点结点
           // head->p1->p2->...  ---->> head->p2->p1->...
           head -> next = p2;
           p1 -> next = p2 -> next;
           p2 -> next = p1;
           //move head to p1
           head = p1;
       }
       return newList->next;
    }
};