java 将文本字符串中的字符移动指定数量的位置
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moving the characters in a text string a specified number of positions
提问by andrey shirokij
I am new in programming and I am trying to write a program that moves the characters in a text string a specified number of positions.
我是编程新手，我正在尝试编写一个程序，将文本字符串中的字符移动指定数量的位置。
The program must include a method whose inputs will be a text string (type String) and the number of positions (type int). The output will be a string with characters shifted.
该程序必须包含一个方法，其输入将是一个文本字符串（类型 String）和位置数（类型int）。输出将是一个字符移位的字符串。
For example, moving 4 positions:
例如，移动 4 个位置：
rabbit eats a carrot it eats a carrotrabb
Now I have this partial code. I can erase first characters but I don't know how to put them to the end of this text. How can i make it?
现在我有了这个部分代码。我可以删除第一个字符，但我不知道如何将它们放在本文的末尾。我怎样才能做到？
public static void main(String[] args) {
System.out.println("enter the text: ");
Scanner cti = new Scanner(System.in);
String a = cti.nextLine();
System.out.println("enter number of positions= ");
int b = cti.nextInt();
char firstLetter = a.charAt(0);
b--;
a = a.substring(b);
String m = a + firstLetter ;
System.out.println("now it is "+ m);
}
采纳答案by bsoren
import java.util.*;
public class JavaApplication5 {
public static void main(String[] args) {
System.out.println("enter the text: ");
Scanner cti = new Scanner(System.in);
String a = cti.nextLine();
System.out.println("enter number of positions= ");
int b = cti.nextInt();
String firstPart = a.substring(0,b); // line 1
b--;
a = a.substring(b);
String m = a + firstPart ; // line 2
System.out.println("now it is "+ m);
}
}See the changes above in statement marked with comment line 1 and line 2.
请参阅上面用注释第 1 行和第 2 行标记的语句中的更改。
In line 1, we are getting the first part of string and in line 2, adding at the end of second string part.
在第 1 行，我们得到字符串的第一部分，在第 2 行，在第二个字符串部分的末尾添加。
回答by sasuke
public String foo(String s, int n) {
String s2 = s.substring(0, n);
s = s.substring(n) + s2;
return s;
}
you can put a few validations on this, like null string or n is less than s.length() etc.
您可以对此进行一些验证，例如空字符串或 n 小于 s.length() 等。
回答by Bohemian
If you use regex, it's just one line:
如果您使用正则表达式，它只是一行：
return str.replaceAll("^(.{" + n + "})(.*)", "");
回答by dnsh
It is better to use modulus operator to calculate number of shifts. When initial number of shift is more than string length. Check this :
最好使用模数运算符来计算移位次数。当初始移位次数大于字符串长度时。检查这个：
public String shift(String string,int n){
int nshift = string.length() < n ? n%string.length() : n ;
String a = string.substring(0,nshift);
return string.substring(nshift) + a ;
}
回答by striving_coder
One more version. All the work is essentially done in 1 line here:
再来一个版本。所有的工作基本上都是在这里的 1 行中完成的：
String result = new StringBuilder(a).delete(0, b).append(a.substring(0,b)).toString();
Anyway, the full code is:
无论如何，完整的代码是：
import java.util.*;
public class ShiftLetters {
public static void main(String[] args) {
System.out.print("enter the text: ");
Scanner cti = new Scanner(System.in);
String a = cti.nextLine();
System.out.print("Enter number of positions: ");
int b = cti.nextInt();
String result = new StringBuilder(a).delete(0, b).append(a.substring(0,b)).toString();
System.out.println(result);
}
}
Also, you might want to be more accurate with your indentation style to improve readability.
此外，您可能希望缩进样式更准确以提高可读性。
