这是什么类型的? [英] What kind of sort is this ?
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问题描述
int x[] = { 123 , 999 , 456 , 3432 , 12 , 43 , 9 , 67 , 9000 , 12345 , 45678 , 789 , 765 , 321 } ;
int x1[] = { 123 , 999 , 456 , 3432 , 12 , 43 , 9 , 67 , 9000 , 12345 , 45678 , 789 , 765 , 321 } ;
for ( int i=0 ; i<=4 ; i++ )
{
int c=0 ;
for ( int k=0 ; k<10 ;k++)
{
for ( int a=0 ; a<x.length ; a++ )
{
if ( ( x1[a] / ( (int)( Math.pow(10,i) ) ) ) %10 == k )
{
x[c] = x1[a] ;
System.out.println(x[c] + " " + c );
c++ ;
}
}
}
for ( int u=0 ; u<x1.length ; u++ )
{
x1[u] = x[u] ;
}
}
我尝试过:
我试图实现基数排序,并在第一次尝试时结束了这一点。我删除了临时数组,因为我仔细检查时不需要它。它是一个好的算法,它是什么类型的?
What I have tried:
I was trying to implement radix sort and ended up with this in first try . I removed the temporary array since it is not required when I checked carefully . Is it a good algorithm and what kind of sort is it ?
推荐答案
学会正确缩进你的代码,它显示它的结构,它有助于阅读和理解。它还有助于发现结构错误。
Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.
int x[] = { 123 , 999 , 456 , 3432 , 12 , 43 , 9 , 67 , 9000 , 12345 , 45678 , 789 , 765 , 321 } ;
int x1[] = { 123 , 999 , 456 , 3432 , 12 , 43 , 9 , 67 , 9000 , 12345 , 45678 , 789 , 765 , 321 } ;
for ( int i=0 ; i<=4 ; i++ )
{
int c=0 ;
for ( int k=0 ; k<10 ;k++)
{
for ( int a=0 ; a<x.length ; a++ )
{
if ( ( x1[a] / ( (int)( Math.pow(10,i) ) ) ) %10 == k )
{
x[c] = x1[a] ;
System.out.println(x[c] + " " + c );
c++ ;
}
}
}
for ( int u=0 ; u<x1.length ; u++ )
{
x1[u] = x[u] ;
}
}
专业程序员的编辑器具有此功能以及其他功能,例如括号匹配和语法高亮。
Notepad ++ Home [ ^ ]
ultraedit [ ^ ]
Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
Notepad++ Home[^]
ultraedit[^]
Quote:
这是一个很好的算法,它是什么类型的?
Is it a good algorithm and what kind of sort is it ?
它看起来像一个基数排序但效率很低。
It look like a radix sort but highly inefficient.
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