顺序表应用5:有序顺序表归并
Time Limit: 100MS Memory limit: 800K
题目描述
输入
第一行输入m、n(1<=m,n<=10000)的值,即为表A、B的元素个数;
第二行输入m个有序的整数,即为表A的每一个元素;
第三行输入n个有序的整数,即为表B的每一个元素;
输出
示例输入
5 3 1 3 5 6 9 2 4 10
示例输出
1 2 3 4 5 6 9 10
#include <stdio.h>
#include <stdlib.h>
#define maxsize 20000
using namespace std;
typedef int element;
typedef struct
{
element *elem;
int length;
int listsize;
} SqList;
int initList(SqList * L)
{
L->elem = new element[maxsize];
if(!L->elem) return -1;
L->length = 0;
L->listsize = maxsize;
return 0;
}
void creat(SqList * L, int n)
{
int i;
for(i=0; i<n; i++)
scanf("%d", &L->elem[i]);
L->length = n;
}
int purge(SqList &La, SqList &Lb, SqList &Lc)
{
if(La.length+Lb.length>Lc.listsize)
return 0;
element *p, *q;
int i=0, j=0, k=0;
while(i<La.length&&j<Lb.length)
{
if(La.elem[i]<Lb.elem[j])
{
Lc.elem[Lc.length++]=La.elem[i];
i++;
}
else
{
Lc.elem[Lc.length++]=Lb.elem[j];
j++;
}
}
while(i<La.length)
{
Lc.elem[Lc.length++]=La.elem[i];
i++;
}
while(j<Lb.length)
{
Lc.elem[Lc.length++]=Lb.elem[j];
j++;
}
for(p=Lc.elem; p<Lc.elem + Lc.length-1; p++)
printf("%d ", *p);
printf("%d", Lc.elem[Lc.length-1]);
}
int main()
{
SqList La;
SqList Lb;
SqList Lc;
int n, i, t, m;
scanf("%d %d", &n, &m);
initList(&La);
creat(&La, n);
initList(&Lb);
creat(&Lb, m);
initList(&Lc);
purge(La, Lb, Lc);
return 0;
}
#include <stdio.h>
#include <stdlib.h>
#define maxsize 20010
typedef int element;
typedef int status;
typedef struct
{
element *elem;
int length;
int listsize;
}sqlist;
status initlist(sqlist &l)
{
l.elem = new element[maxsize];
if(!l.elem) exit(-1);
l.length = 0;
l.listsize = maxsize;
return 0;
}
status create(sqlist &l, int n)
{
int i;
for(i=0; i<n; i++)
{
scanf("%d", &l.elem[i]);
}
l.length = n;
return 0;
}
void Merge(sqlist &la, sqlist &lb, sqlist &lc)
{
int i=0, j=0;
while(i<la.length&&j<lb.length)
{
if(la.elem[i]<lb.elem[j])
{
lc.elem[lc.length++] = la.elem[i++];
}
else
{
lc.elem[lc.length++] = lb.elem[j++];
}
}
while(i<la.length)
{
lc.elem[lc.length++] = la.elem[i++];
}
while(j<lb.length)
{
lc.elem[lc.length++] = lb.elem[j++];
}
}
void display(sqlist &l)
{
int i;
for(i=0; i<l.length-1; i++)
{
printf("%d ", l.elem[i]);
}
printf("%d\n", l.elem[l.length-1]);
}
int main()
{
int n, m;
sqlist la, lb, lc;
scanf("%d %d", &n, &m);
initlist(la);
initlist(lb);
initlist(lc);
create(la, n);
create(lb, m);
Merge(la, lb, lc);
display(lc);
return 0;
}