/* package codechef; // don't place package name! */
import java.util.StringTokenizer;
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
String func (String s1)
{
StringTokenizer st =new StringTokenizer(s1);
String fina= new String();
while(st.hasMoreTokens())
{
String s= st.nextToken();
char []ch= s.toCharArray();
int length= ch.length;
int r=0;
int i =1;
while(i<=length/2)
{
char temp;
temp=ch[i];
ch[i]=ch[length-1-r];
ch[length-1-r]=temp;
r++;
i++;
}
String revword=new String(ch);
fina+=""+revword +" ";
}
return(fina);
}
public static void main (String[] args) throws java.lang.Exception
{
String s1="Tarun is a intern";
Codechef c=new Codechef();
String s2=c.func(s1);
System.out.println(""+s2);
}
}
The above code was run on codechef ide i am using 2638 KB of memory. Is there any other method I can use to decrease the memory usage? Or any other optimized method? I tried using StringBuilder to append as it is mutable still it was giving me same memory usage on code chef.
上面的代码是在codechef上运行的,我正在使用2638 KB的内存。有没有其他方法可以用来减少内存使用量?还是其他任何优化方法?我尝试使用StringBuilder追加,因为它是可变的仍然它给了我相同的内存使用代码厨师。
1 个解决方案
#1
1
This question would probably be better on codereview.stackexchange.com but nevertheless, I'll make some comments here.
这个问题在codereview.stackexchange.com上可能会更好,但不过,我会在这里发表一些评论。
Out of the 2638 KB of memory used, I expect the vast majority of that is used by the Java Runtime Environment, rather than directly by your application. Still you might make a few KB of savings. Here's a line-by-line commentary.
在使用的2638 KB内存中,我预计绝大多数内存都是由Java Runtime Environment使用的,而不是由您的应用程序直接使用。你仍然可以节省几KB。这是一个逐行评论。
/* package codechef; // don't place package name! */
No need to keep the above line in your code.
无需在代码中保留上述行。
import java.util.StringTokenizer;
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
String func (String s1)
{
StringTokenizer st =new StringTokenizer(s1);
The documentation of StringTokenizer says this is a legacy class kept for backward compatibility and that newer code should use String.split() instead. This would improve readability, though I doubt it would improve performance or memory usage.
StringTokenizer的文档说这是为了向后兼容而保留的遗留类,而较新的代码应该使用String.split()。这会提高可读性,但我怀疑它会提高性能或内存使用率。
String fina= new String();
Why are you using new String()? Just use "". By using new String() you are creating a new object, thus wasting a few bytes of memory, when "" refers to an existing object, and is more readable.
你为什么使用新的String()?只需使用“”。通过使用新的String(),您将创建一个新对象,从而浪费几个字节的内存,当“”引用现有对象时,并且更具可读性。
while(st.hasMoreTokens())
{
String s= st.nextToken();
char []ch= s.toCharArray();
int length= ch.length;
int r=0;
It may be slightly more readable to start r at 1 and use length-r instead of length-1-r.
从1开始r并使用length-r而不是length-1-r可能稍微更具可读性。
int i =1;
You need to start i at 0, as this is the index of the first character.
你需要从0开始,因为这是第一个字符的索引。
while(i<=length/2)
It may be more readable to use a for loop.
使用for循环可能更具可读性。
{
char temp;
temp=ch[i];
Why not declare and initialize on the same line? char temp = ch[i];
为什么不在同一行上声明和初始化? char temp = ch [i];
ch[i]=ch[length-1-r];
ch[length-1-r]=temp;
r++;
i++;
}
String revword=new String(ch);
fina+=""+revword +" ";
There is no need for the ""+ preceding revword. You are creating a new String by doing this (thus wasting memory). But revword is already a string, so no need to append it to a null string.
不需要“+”前面的revword。您正在通过执行此操作来创建新的String(从而浪费内存)。但是revword已经是一个字符串,所以不需要将它附加到空字符串。
Appending to a String fina inside a loop costs CPU time (and some memory). You should use a StringBuilder instead.
在循环内附加到字符串fina会花费CPU时间(和一些内存)。您应该使用StringBuilder。
Also, if you used StringBuilder you would have been able to copy ch to it directly, without having to create a new String that is a copy of ch.
此外,如果您使用StringBuilder,您可以直接将ch复制到它,而无需创建一个新的String作为ch的副本。
}
return(fina);
Minor nit-pick: no need for parentheses here, as it makes the return keyword look like a method call, which it is not.
次要的挑选:这里不需要括号,因为它使return关键字看起来像方法调用,但它不是。
}
By the way, your method adds a " " (space) at the end of the returned string, which is probably not what you want.
顺便说一句,你的方法在返回的字符串的末尾添加一个“”(空格),这可能不是你想要的。
public static void main (String[] args) throws java.lang.Exception
{
String s1="Tarun is a intern";
Codechef c=new Codechef();
String s2=c.func(s1);
System.out.println(""+s2);
}
}
#1
1
This question would probably be better on codereview.stackexchange.com but nevertheless, I'll make some comments here.
这个问题在codereview.stackexchange.com上可能会更好,但不过,我会在这里发表一些评论。
Out of the 2638 KB of memory used, I expect the vast majority of that is used by the Java Runtime Environment, rather than directly by your application. Still you might make a few KB of savings. Here's a line-by-line commentary.
在使用的2638 KB内存中,我预计绝大多数内存都是由Java Runtime Environment使用的,而不是由您的应用程序直接使用。你仍然可以节省几KB。这是一个逐行评论。
/* package codechef; // don't place package name! */
No need to keep the above line in your code.
无需在代码中保留上述行。
import java.util.StringTokenizer;
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
String func (String s1)
{
StringTokenizer st =new StringTokenizer(s1);
The documentation of StringTokenizer says this is a legacy class kept for backward compatibility and that newer code should use String.split() instead. This would improve readability, though I doubt it would improve performance or memory usage.
StringTokenizer的文档说这是为了向后兼容而保留的遗留类,而较新的代码应该使用String.split()。这会提高可读性,但我怀疑它会提高性能或内存使用率。
String fina= new String();
Why are you using new String()? Just use "". By using new String() you are creating a new object, thus wasting a few bytes of memory, when "" refers to an existing object, and is more readable.
你为什么使用新的String()?只需使用“”。通过使用新的String(),您将创建一个新对象,从而浪费几个字节的内存,当“”引用现有对象时,并且更具可读性。
while(st.hasMoreTokens())
{
String s= st.nextToken();
char []ch= s.toCharArray();
int length= ch.length;
int r=0;
It may be slightly more readable to start r at 1 and use length-r instead of length-1-r.
从1开始r并使用length-r而不是length-1-r可能稍微更具可读性。
int i =1;
You need to start i at 0, as this is the index of the first character.
你需要从0开始,因为这是第一个字符的索引。
while(i<=length/2)
It may be more readable to use a for loop.
使用for循环可能更具可读性。
{
char temp;
temp=ch[i];
Why not declare and initialize on the same line? char temp = ch[i];
为什么不在同一行上声明和初始化? char temp = ch [i];
ch[i]=ch[length-1-r];
ch[length-1-r]=temp;
r++;
i++;
}
String revword=new String(ch);
fina+=""+revword +" ";
There is no need for the ""+ preceding revword. You are creating a new String by doing this (thus wasting memory). But revword is already a string, so no need to append it to a null string.
不需要“+”前面的revword。您正在通过执行此操作来创建新的String(从而浪费内存)。但是revword已经是一个字符串,所以不需要将它附加到空字符串。
Appending to a String fina inside a loop costs CPU time (and some memory). You should use a StringBuilder instead.
在循环内附加到字符串fina会花费CPU时间(和一些内存)。您应该使用StringBuilder。
Also, if you used StringBuilder you would have been able to copy ch to it directly, without having to create a new String that is a copy of ch.
此外,如果您使用StringBuilder,您可以直接将ch复制到它,而无需创建一个新的String作为ch的副本。
}
return(fina);
Minor nit-pick: no need for parentheses here, as it makes the return keyword look like a method call, which it is not.
次要的挑选:这里不需要括号,因为它使return关键字看起来像方法调用,但它不是。
}
By the way, your method adds a " " (space) at the end of the returned string, which is probably not what you want.
顺便说一句,你的方法在返回的字符串的末尾添加一个“”(空格),这可能不是你想要的。
public static void main (String[] args) throws java.lang.Exception
{
String s1="Tarun is a intern";
Codechef c=new Codechef();
String s2=c.func(s1);
System.out.println(""+s2);
}
}