1.判断一个链表是否存在环，例如下面这个链表就存在一个环：例如N1->N2->N3->N4->N5->N2就是一个有环的链表

c语言版：

1 #include 
      
        2 #include 
       
         3  4 struct link{ 5     int data; 6     struct link *next; 7 }; 8  9 int isLoop(struct link* head){10     struct link* p1,*p2;11     if(head == NULL || head->next == NULL)12         return 0;13     p1 = head;14     p2 = head;15     do{16         p1 = p1->next;17         p2 = p2->next->next;18     }while(p1 != NULL && p2 != NULL && p1 != p2);19     if(p1 == p2)20         return 1;21     else22         return 0;23 24 }25 26 int main()27 {28     29 }
       
      

 

2.单向链表的反转

c语言版：

1 #include 
     
       2 #include 
      
        3  4 struct link{ 5     int data; 6     struct link *next; 7 }; 8  9 void reserve(struct link* head){10     if(head == NULL || head->next == NULL)11         return;12     struct link * pre,*nex,*p;13     pre = head;14     p = head->next;15     nex = head->next->next;16     do{17         p->next = pre;18         pre->next = NULL;19         pre = p;20         p = nex;21         nex = nex->next;22     }while(nex != NULL);23     head = p;24 25 }26 27 int main()28 {29 30 }
      
     

c语言（递归法）：

1 #include 
     
       2 #include 
      
        3  4 struct link{ 5     int data; 6     struct link *next; 7 }; 8 //递归法 9 struct link* reserve(struct link* p,struct link* head){10     if(p == NULL || p->next == NULL){11         head = p;12     }13     else{14         struct link* temp = reserve(p->next,head);15         temp->next = p;16         return p;17     }18 }19 20 int main()21 {22 23 }