~~~How can the writing style end in a dull way, and the story does not recognize ordinary at the beginning ✌✌✌
If you need the complete code, you can follow the public account below, and reply "code" in the background to get it. Aguang is looking forward to your visit~
Topic:
Let C={a1,b1,a2,b2} be a linear table, use the hc singly linked list with the head node to store, design an in-place algorithm, split it into two linear tables, so that A={a1,a2 ,an}, B={bn,b2,b1}.
Problem solving ideas:
>问题的关键就是采用何种方式构建链表
>A采用尾插保持原顺序
>B采用前插法将其逆序
Code:
#include <iostream>
using namespace std;
typedef struct LNode
{
int data;
struct LNode *next;
} LNode, *LinkList;
// 头插法
void HeadInsert(LinkList &L)
{
int val = 0;
while (cin >> val)
{
LNode *s = new LNode;
s->data = val;
s->next = L->next;
L->next = s;
if (cin.get() == '\n')
{
break;
}
}
}
// 尾插法
void TailInsert(LinkList &L)
{
int val = 0;
LNode *r = L;
while (cin >> val)
{
LNode *s = new LNode;
s->data = val;
r->next = s;
r = s;
r->next = NULL;
if (cin.get() == '\n')
{
break;
}
}
}
// 遍历输出链表元素
void Print(LinkList L)
{
LNode *p = L->next;
while (p)
{
cout << p->data << '\t';
p = p->next;
}
cout << endl;
}
void BreakList(LinkList &LA, LinkList &LB)
{
LNode *p, *q, *ra; //工作指针、保存后继指针、LA尾指针
LB->next = NULL; //将LB链表置空
p = LA->next;
ra = LA; //初始化尾指针
while (p)
{
// LA尾插
ra->next = p;
ra = p;
p = p->next;
//如果p不为空,继续将其前插到B
if (p)
{
q = p->next; //保存p的后继,防止断链
p->next = LB->next;
LB->next = p;
p = q;
}
}
ra->next = NULL; //将A的尾结点的指针域置为空
}
int main()
{
LinkList LA = new LNode;
LinkList LB = new LNode;
TailInsert(LA);
BreakList(LA, LB);
Print(LA);
Print(LB);
}