数据结构第二章课后作业

设L的型为LIST线性表实例，x 的型为elementtype的元素

            实例，p 为位置变量。所有操作描述为：

            ①  INSERT(x, p, L)

            ②  LOCATE(x, L)

            ③  RETRIEVE(p, L)

            ④  DELETE(p, L)

            ⑤  PREVIOUS(p, L),  NEXT(p, L)

            ⑥  MAKENULL( L)

            ⑦  FIRST( L)                                                      ⑧END(L)

1.写一个算法合并两个已排序的线性表

伪代码：

定义一个函数： position  Attach( int c ,position d)   //建立一个新节点，并把它链接到d所指的节点之后，并返回这个新节点的指针

position linkadd(position A ,position B)
{
    position p,q,d,c;
    Elymenttype x ;
    c = new node ; d = c;
    while (p!=END(a)&&q!=END(b))
    {
        if(RETRIEVE(p, A)==RETRIEVE(q, B))
        {
            d =Attach(RETRIEVE(p, A),d ) //将新节点链入
            p=NEXT(p,A);
            q =NEXT(q,B);
        }
        else if (RETRIEVE(p, A)>RETRIEVE(q, B))
        {
             d =Attach(RETRIEVE(p, A),d ) //将新节点链入
             p=NEXT(p,A);
        }
        else 
        {
             d =Attach(RETRIEVE(q, B),d ) //将新节点链入
             q=NEXT(q,B);
        }
    }
    while(p!=END(A))
    {
        d =Attach(RETRIEVE(p, A),d ) //将新节点链入
        p=NEXT(p,A);
    }
    while(p!=END(B))
    {
        d =Attach(RETRIEVE(q, B),d ) //将新节点链入
        p=NEXT(q,B);
    }
}

从C/C++

#include <iostream>  
#include<stdio.h>  
#include<stdlib.h>  
using namespace std;  
struct polynode {  
    int exp;  
    polynode *next;  
};  
polynode *createapolylink()  
{  
    int n, i, exp;  
    polynode *head, *pre, *p;  
    printf("输入链表长度
");  
    scanf_s("%d", &n);  
    head = new polynode;  
    head->next = NULL;  
    pre = head;  
  
    for (i = 0; i<n; i++)  
    {  
        printf("链表第%d个数", i + 1);  
        p = new polynode;  
        scanf_s(" %d",  &exp);  
        p->exp = exp;  
        p->next = NULL;  
        pre->next = p;  
        pre = p;  
  
    }  
    return head;  
}  
void printfthepolylink(polynode *head)  
{  
    polynode *p;  
    p = head->next;  
    while (p != NULL)  
    {  
        printf("%d ",  p->exp);  
        p = p->next;  
    }  
}  
polynode *attach(int e, polynode *d)  
{  
    polynode *x;  
    x = new polynode;  
    x->exp = e;  
    d->next = x;  
    return x;  
}  
polynode *polyadd(polynode* a, polynode *b)  
{  
    polynode * p, *q, *d, *c;  
    p = a;  
    q = b;  
    c = new polynode; d = c;  
    while ((p != NULL) && (q != NULL))  
    {  
        if (p->exp == q->exp)  
        {  
            d = attach(p->exp, d);  
            p = p->next; q = q->next;  
        }  
        else if (p->exp>q->exp)  
        {  
            d = attach( p->exp, d);  
            p = p->next;  
        }  
        else  
        {  
            d = attach( q->exp, d);  
            q = q->next;  
        }  
    }  
    while (p != NULL)  
    {  
        d = attach( p->exp, d);  
        p = p->next;  
    }  
    while (q != NULL)  
    {  
        d = attach( q->exp, d);  
        q = q->next;  
    }  
    d->next = NULL;  
    p = c; c = c->next; delete p;  
    return c;  
}  
int main()  
{  
    polynode *link1, *link2, *link3;  
    link1 = createapolylink();  
    link2 = createapolylink();  
    link3 = polyadd(link1, link2);  
    printfthepolylink(link3);  
    system("pause");  
    return 0;  
}  

2.已知一个字符串，次序为3*-y -a/y ↑2，试利用栈把该字符串的次序变为3y-*ay2↑/-的操作步骤