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法律研究基於引用區域性性的 C++ 搜尋程式
基於引用區域性性的搜尋依賴於記憶體訪問模式，其中資料元素被重新分配。
此處使用線性搜尋方法來搜尋元素。
演算法
Begin int find(int *intarray, int n, int item) intialize comparisons = 0 for i = 0 to n-1 Increase comparisons if(item == intarray[i]) Print element with its index break if(i == n-1) Print element not found return -1 Print Total comparisons For j = i till i>0 intarray[j] = intarray[j-1] intarray[0] = item return 0 End
示例
#include<iostream>
using namespace std;
// A function to perform linear search.
// this method makes a linear search.
int find(int *intarray, int n, int item) {
int i;
int comparisons = 0;
// navigate through all items
for(i = 0;i<n;i++) {
// count the comparisons made comparisons++;
// if item found, break the loop
if(item == intarray[i]) {
cout<<"element found at:"<<i<<endl;
break;
}
// If index reaches to the end then the item is not there.
if(i == n-1) {
cout<<"\nThe element not found.";
return -1;
}
}
printf("Total comparisons made: %d", comparisons);
// Shift each element before matched item.
for(int j = i; j > 0; j--)
intarray[j] = intarray[j-1];
// Put the recently searched item at the beginning of the data array.
intarray[0] = item;
return 0;
}
int main() {
int intarray[20]={1,2,3,4,6,7,9,11,12,14,15,16,26,19,33,34,43,45,55,66};
int i,n;
char ch;
// print initial data array.
cout<<"\nThe array is: ";
for(i = 0; i < 20;i++)
cout<<intarray[i]<<" ";
up:
cout<<"\nEnter the Element to be searched: ";
cin>>n;
// Print the updated data array.
if(find(intarray,20, n) != -1) {
cout<<"\nThe array after searching is: ";
for(i = 0; i <20;i++)
cout<<intarray[i]<<" ";
}
cout<<"\n\nWant to search more.......yes/no(y/n)?";
cin>>ch;
if(ch == 'y' || ch == 'Y')
goto up;
return 0;
}輸出
The array is: 1 2 3 4 6 7 9 11 12 14 15 16 26 19 33 34 43 45 55 66 Enter the Element to be searched: 26 element found at:12 Total comparisons made: 13 The array after searching is: 26 1 2 3 4 6 7 9 11 12 14 15 16 19 33 34 43 45 55 66 Want to search more.......yes/no(y/n)?y Enter the Element to be searched: 0 The element not found. Want to search more.......yes/no(y/n)?n
更新日期：30-Jul-2019
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