संदर्भ के स्थान के आधार पर खोजना, मेमोरी एक्सेस पैटर्न पर निर्भर करता है डेटा तत्वों को पुनः आवंटित किया जाता है।
यहां किसी तत्व को खोजने के लिए रैखिक खोज विधि का उपयोग किया जाता है।
एल्गोरिदम
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 Endके लिए
उदाहरण
#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;
} आउटपुट
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