您的位置:首页 > 其它

线性表--算法设计题2.23

2011-04-28 01:42 323 查看
设线性表A=(a1,a2……,am),B=(b1,b2……,bn),试写一个按下列规则合并A,B为线性表C的算法,即使得
C=(a1,b1,……am,bm,……bn) 当m<=n时;

C=(a1,b1,……an,bn,……am) 当m>n时;

C code:

#include<stdio.h>

#include<stdlib.h>
#define LIST_INIT_SIZE 10
#define LIST_INCREMENT 2
#define ERROR 0
#define OK 1
#define OVERFLOW -1
#define TRUE 1
typedef int Status;

struct SqList
{
int *elem;
int length;
int listsize;
};

void InitList(SqList &L)
{
L.elem=(int*)malloc(LIST_INIT_SIZE*sizeof(int));
if(!L.elem)
exit(OVERFLOW);
L.length=0;
L.listsize=LIST_INIT_SIZE;
}

Status ListInsert(SqList &L,int i,int e)
{
int *newbase,*q,*p;
if(i<1||i>L.length+1)
return ERROR;
if(L.length==L.listsize)
{
newbase=(int*)realloc(L.elem,(L.listsize+LIST_INCREMENT)*sizeof(int));
if(!newbase)
exit(OVERFLOW);
L.elem=newbase;
L.listsize+=LIST_INCREMENT;

}
q=L.elem+i-1;
for(p=L.elem+L.length-1; p>=q; --p)
*(p+1)=*p;
*q=e;
++L.length;
return OK;
}

void CreateList(SqList &L,int len)
{
int i;
for(i=1; i<=len; i++)
{
ListInsert(L,i,rand()%100);
}
}

void CreateList_C(SqList La,SqList Lb,SqList &Lc)
{
int i,j;
if(Lc.listsize<=La.length+Lb.length)
{
Lc.listsize+=La.length+Lb.length;
}
for(i=0,j=1; j<=La.length+Lb.length && i<La.length && i<Lb.length; i++)
{
ListInsert(Lc,j,La.elem[i]);
ListInsert(Lc,j+1,Lb.elem[i]);
j+=2;
}
if(La.length<Lb.length)
{
for(; i<Lb.length; i++,j++)
{
ListInsert(Lc,j,Lb.elem[i]);
}
}
else
{
for(; i<La.length; i++,j++)
{
ListInsert(Lc,j,La.elem[i]);
}
}
}

void ListTraverse(SqList L,void(*vist)(int&))
{
int *p=L.elem;
int i;
for(i=1; i<=L.length; i++)
{
vist(*p++);
}
printf("\n");
}

void print1(int &c)
{
printf("%d ",c);
}

int main()
{
SqList La,Lb,Lc;
InitList(La);
InitList(Lb);
InitList(Lc);
int La_len,Lb_len;
printf("input length of La: ");
scanf("%d",&La_len);
printf("input length of Lb: ");
scanf("%d",&Lb_len);
CreateList(La,La_len);
CreateList(Lb,Lb_len);
printf("List a: ");
ListTraverse(La,print1);
printf("List b: ");
ListTraverse(Lb,print1);
CreateList_C(La,Lb,Lc);
printf("List c: ");
ListTraverse(Lc,print1);
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: 
相关文章推荐