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 <stdio.h> 2 #include <stdlib.h> 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 else 22 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 <stdio.h> 2 #include <stdlib.h> 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 <stdio.h> 2 #include <stdlib.h> 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 }