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;    }};