分块查找，代码已写好了，无错误，运行时报内存错误
分块查找，代码已写好了，无错误，运行时报内存错误
题是在8,14,6,9,10,22,34,18,19,31,40,38,54,66,46,71,78,68,80,85,100,94,88,96,87中找46的位置
代码如下
#include<stdio.h>
#define MAXL 100
#define MAXI 20
typedef int KeyType;
typedef char InfoType[10];
typedef struct
{
KeyType key;
InfoType data;
}NodeType;
typedef NodeType SeqList[MAXL];
typedef struct
{
KeyType key;
int link;
}IdxType;
typedef IdxType IDX[MAXI];
int IdxSearch(IDX I,int m,SeqList R,int n,KeyType k)
{
int low=0,high=m-1,mid,i,count1=0,count2=0;
int b=n/m;
printf("二分查找\n");
while(low<=high)
{
mid=(low+high)/2;
printf("第%n次查找:在[%d,%d]中查找到元素R[%d]:%d\n",count1+1,low,high,mid,R[mid].key);
if(I[mid].key>=k)
high=mid-1;
else
low=mid+1;
count1++;
}
if(low<m)
{
printf("比较%d次,在第%d块中查找元素%d\n",count1,low,k);
i=I[low].link;
printf("顺序查找:\n");
while(i<=I[low].link+b-1&&R[i].key!=k)
{
i++;
count2++;
printf("%d",R[i].key);
}
printf("\n");
printf("比较%d次,在顺序表中查找元素%d\n",count2,k);
if(i<=I[low].link+b-1)
return i;
else
return -1;
}
return -1;
}
void main()
{
SeqList R;
KeyType k=46;
IDX I;
int a[]={8,14,6,9,10,22,34,18,19,31,40,38,54,66,46,71,78,68,80,85,100,94,88,96,87},i;
for(i=0;i<25;i++)
R[i].key=a[i];
I[0].key=14;I[0].link=0;
I[1].key=34;I[1].link=4;
I[2].key=66;I[2].link=10;
I[3].key=85;I[3].link=15;
I[4].key=100;I[4].link=20;
printf("\n");
if((i=IdxSearch(I,5,R,25,k))!=-1)
printf("\n元素%d的位置是%d\n",k,i);
else
printf("\n元素%d不在表中\n",k);
printf("\n");
}
[解决办法]
如果数据量不大，这种问题很好解决啊，不用分块的，扫描一次数组，直到找到这个值，记录下标，退出即可。
如果要找出所有值，就全部遍历
