LeetCode_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

题目：把给定链表中的相邻的两个节点交换翻转，并返回；

例如：1->2->3->4 输出： 2->1->4->3；

注意：涉及到指针的操作，注意前两个节点在交换的时候要注意head指针的变化。

代码：

/**
 * 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;
       ListNode *p = head;
       ListNode *q = p->next;  
       if(q==NULL) return head;
       int flag = 0;
       while(q!=NULL&&p!=NULL)
       {
              if(flag == 0)//前两个节点交换
              head = q;
              
              p->next = q->next;
              q->next = p;
              
              ListNode *p1 = p;
              
              p=p->next;
              if(p==NULL)
              break;
              else
              {
                 q=p->next;
                 if(q!=NULL)
                 p1->next = q;
                 flag = 1;
              }
       }
       return head;
    }
};