删除链表中重复的数

题目：

给定一个排序的链表，将其中重复的数全部删除。

比如：1-->1-->1-->2-->3，则返回2-->3

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

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

ListNode* rmDumplicateNums(ListNode* head)
{
   ListNode stHead = {0};
   ListNode* node = head;
   ListNode* p = &stHead;
   ListNode* next;
   ListNode* tmp;
   bool flag;

   /*此题采用尾插法即可*/
   while (node)
   {
       flag = false;
       next = node->next;
       /*找到相等的结点*/
       while (next && (node->val==next->val))
       {
            next = next->next;
            flag = true;
       }

       /*有就删除*/
       if (flag)
       {
           while (node != next)
           {
               tmp = node->next;
               free(node);
               node = tmp;
           }
       }
       /*否则插入*/
       else
       {
           p->next = node;
           p = p->next;
       }

       node = next;
   }
   p->next = NULL;

   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, 5, 1, 1, 1};
     createList(&stHead, a, 5);
     printList(&stHead);
     stHead.next = rmDumplicateNums(stHead.next);
     printList(&stHead);
     return 0;
}