两两交换链表的结点

题目：

例如链表为1-->2-->3-->4，则交换后为:2-->1-->4-->3

代码：

#include <stdio.h>
#include <stdlib.h>

typedef struct ListNode
{
    ListNode* next;
    int val;
};

ListNode* swapNodeInPair(ListNode* head)
{
   /*思路:直接交换*/
   ListNode stHead = {0};
   ListNode* pre;
   ListNode* cur;
   ListNode* next;

   stHead.next = head;
   pre = &stHead;
   cur = stHead.next;

   while (cur->next)
   {
       next = cur->next;
       cur->next = next->next;
       pre->next = next;
       next->next = cur;
       pre = cur;
       cur = cur->next; /*这里要注意，由于交换后，其实就跳过两个结点*/
   }

   return stHead.next;
}

void createList(ListNode* head, int* num, int n)
{
     ListNode* p;
     int i;

     for (i=0; i<n; i++)
     {
        p = (ListNode*)malloc(sizeof(ListNode));
        p->next = head->next;
        p->val = num[i];
        head->next = p;
     }
}

void printList(ListNode* head)
{
   ListNode* p = head->next;

   while (p)
   {
        printf("%d ", p->val);
        p = p->next;
   }

   printf("\n");
}

int main()
{
     ListNode stHead = {0};
     int a[] = {5, 4, 3, 2, 1};
     createList(&stHead, a, 5);
     printList(&stHead);
     stHead.next = swapNodeInPair(stHead.next);
     printList(&stHead);
     return 0;
}