Reputation: 5648
public class Zigzag{
public static void zigzag_optimizated(int n, int m) {
int length = 2*m;
int localIndex[] = new int[n];
for(int i = 0; i<n; i++){
localIndex[i] = i % length;
}
for (int i = 0; i <= m; i++) {
for (int j = 0; j < n; j++) {
if (localIndex[j]==i || localIndex[j] == length-i)
assert true;
// System.out.print('*');
else
assert true;
//System.out.print('-');
}
//System.out.println();
assert true;
}
}
public static void zigzag(int n, int m) {
for (int i = 0; i <= m; i++) {
for (int j = 0; j < n; j++) {
int k = j % (2*m);
char c = '-';
if (k==i || k == 2*m-i) c = '*';
assert true;
//System.out.print(c);
}
assert true;
//System.out.println();
}
}
public static void main(String args[]){
final int n = 5000000;
long start = System.nanoTime();
zigzag(n, n);
long time = System.nanoTime() - start;
long start2 = System.nanoTime();
zigzag_optimizated(n, n);
long time2 = System.nanoTime() - start2;
System.out.println();
System.out.println("Time1:" + time);
System.out.println("Time2:" + time2);
}
}
Two functions have same algorithm, it print a zigzag board to screen.
In optimizated version, k is saved in array to avoid recalculate, 2*m is extracted.
I changed System.out.println()
to assert true;
for faster and more accurate benchmark, but when i do the benchmark, the original version is always run faster (with n large enough)
Upvotes: 0
Views: 82
Reputation: 99
How big is n to see the difference?
If n is big enough array is too big to keep it in CPU cache - it's faster to calculate j % (2*m) then access it from RAM (60-100 nanosec).
See Scott Mayers - How CPU Cache works and why you care -
Upvotes: 1