带头结点和不带头结点的单连表的第一个元素分别是怎么取得的，最好写出算法……

带头结点和不带头结点的单连表的第一个元素分别是怎么取得的，最好写出算法……

假设 struct node{
int data;
struct node *nexe;}
带头结点时：第一个元素就是 L->next->data;
不带头结点时；第一个元素就是 L->data;
（其中L是指向第一个节点的指针）

带头结点 第一个元素就是L->next 不带头结点 第一个元素就是L

举手之劳，帮你弄了。程序就是算法，既然是顺序表，就按顺序查找就可以了。你要用二分法查找？那也可以啊。明天来写。 程序的数据可以根据实际情况更改，注意程序中的人名第一个字母这里为大写。一个完整的c程序如下，在win-tc和dev-c++下已运行通过。 #include #include #include #include #define num 4 struct chain { char name[20]; char city[20]; char sex[10]; char age[10]; char job[10]; struct chain *next; }; struct chain *create(); struct chain *sequelseach(struct chain *head,char name[]); void print_data(struct chain *point); struct chain datas[num]= { "sun","wuhan","male","24","student",null, "tom","beijing","male","31","doctor",null, "marry","shanghai","female","19","teacher",null, "willing","tianjing","female","21","worker",null }; int main(void) { struct chain *head; struct chain *p; char name[30]; head=create(); printf("please input the name:\n"); scanf("%s",name); p=sequelseach(head,name); print_data(p); getch(); return 0; } struct chain *create() { struct chain *head, *tail, *p; int i; head=tail=null; printf("put the datas into the list.\n"); for(i= 0;i < num; i++) { p=(struct chain *)malloc (sizeof (struct chain)); strcpy(p->name, datas[i].name); strcpy(p->city,datas[i].city); strcpy(p->sex,datas[i].sex); strcpy(p->age,datas[i].age); strcpy(p->job,datas[i].job); p->next=null; if(head==null) head=tail=p; else tail=tail->next; tail->next=p; } return head; } struct chain *sequelseach(struct chain *head, char name[]) { struct chain *temp; temp=head; for(temp=head;temp!=null;) { if(strcmp(temp->name,name)==0) break; else temp=temp->next; } if(temp ==null) printf("no found!\n"); return temp; } void print_data(point) struct chain *point; { if(point==null) return; printf("results:\n"); printf(" name:%s\n",point->name); printf(" city:%s\n",point->city); printf(" sex:%s\n",point->sex); printf(" age:%s\n",point->age); printf(" profession:%s\n",point->job); }