单链的逆置
摘要:
node *reverse(node *head){node *p1,*p2,*p3;if(head==NULL||head->next==NULL)return head;p1=head,p2=p1->next;while(p2){p3=p2->next;p2->next=p1;p1=p2;p2=p3;}head->next=NULL;head=p1;return head;}int main(){node *head,stud;int n,del,num,insert_num;head=create();print(head);cout<<&quo 阅读全文
posted @ 2013-02-23 12:10 峻华 阅读(110) 评论(0) 推荐(0)
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