I want to calculate the last 20 leap year for which I have written this code, can this code be further optimized

huangapple
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英文:

I want to calculate the last 20 leap year for which I have written this code, can this code be further optimized

问题

  1. package javaHM;
  3. import java.util.*;
  5. public class LeapYear {
  7.     public static void main(String[] args) {
  8.         Scanner obj = new Scanner(System.in);
  9.         System.out.println("Enter Starting Year");
  10.         int year = obj.nextInt();
  11.         obj.close();
  12.         leapFor(year);
  13.     }
  15.     public static void leapFor(int year) {
  16.         int counter = 0;
  17.         for (int i = year; counter < 20; i--) {
  18.             if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0) {
  19.                 System.out.println(i);
  20.                 counter++;
  21.             }
  22.         }
  23.     }
  24. }
英文: 
  1. package javaHM;
  3. import java.util.*;
  5. public class LeapYear {
  7. 	public static void main(String[] args) {
  8. 		// TODO Auto-generated method stub
  9. 		Scanner obj = new Scanner(System.in);
  10. 		System.out.println(&quot;Enter Starting Year&quot;);
  11. 		int year = obj.nextInt();
  12. 			leapFor(year);
  13. 			
  14. 		obj.close();
  16. 	}
  18. 	public static int leapFor(int year) {
  19. 		int counter = 0, i;
  20. 		for (i = year; i &gt;= 1; i--) {
  21. 			if ((i % 4 == 0) &amp;&amp; i % 100 != 0 || i % 400 == 0) {
  22. 				System.out.println(i);
  23. 				counter++;
  25. 			}
  26. 			if (counter == 20)
  27. 				break;
  28. 		}
  29. 		return i;
  31. 	}
  32. }

I want to optimize this portion of code , please help me

答案1

得分: 0

您的函数在每次迭代中执行I/O操作。I/O操作速度较慢，因此具有最大影响的优化是将输出累积到缓冲区中，然后仅在最后一次打印。对算法的任何更改都不会产生有意义的差异。

  1. public static int leapFor(int year) {
  2.     int counter = 0, i;
  3.     StringBuilder output = new StringBuilder(20*(4+1));
  4.     for (i = year; i >= 1; i--) {
  5.         if ((i % 4 == 0) && i % 100 != 0 || i % 400 == 0) {
  6.             output.append(i).append('\n');
  7.             counter++;
  8.         }
  9.         if (counter == 20)
  10.             break;
  11.     }
  12.     System.out.print(output);
  13.     return i;
  14. }
英文:

Your function does I/O on every iteration. I/O is slow, so the optimization with the biggest effect is accumulating the output into a buffer, and printing it only once. Any changes to the arithmetic will not make a meaningful difference.

  1. public static int leapFor(int year) {
  2.     int counter = 0, i;
  3.     StringBuilder output = new StringBuilder(20*(4+1));
  4.     for (i = year; i &gt;= 1; i--) {
  5.         if ((i % 4 == 0) &amp;&amp; i % 100 != 0 || i % 400 == 0) {
  6.             output.append(i).append(&#39;\n&#39;);
  7.             counter++;
  9.         }
  10.         if (counter == 20)
  11.             break;
  12.     }
  13.     System.out.print(output);
  14.     return i;
  15. }

huangapple
  • 本文由 发表于 2020年8月7日 01:17:04
  • 转载请务必保留本文链接：https://go.coder-hub.com/63288692.html
  • java
  • optimization
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