Java- there are n integers, each front so that the sequence number m set back position, and finally into the number m of the number m of the foremost

<How a>

package number;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Number {

	public static void main(String[] args) {
		
		List<String> list=new LinkedList<String>();
		Scanner input=new Scanner(System.in);
		
		System.out.println("请输入一串数字：");
		String str=input.nextLine();
		System.out.println("请输入后移位置的m个数：");
		int m=input.nextInt();
		
		for (int i = str.length()-1; i >=0; i--) {
			
			if(i>=m) {
				((LinkedList<String>) list).addFirst(str.substring(i, i+1));
			}else {
				((LinkedList<String>) list).add(str.length()-m,str.substring(i, i+1));
			}
		}
		for (int i = 0; i < list.size(); i++) {
			System.out.print(list.get(i));
		}
		
	}
}

<Practice two>

import java.util.Scanner;

public class NumberDemo {

	public static void main(String[] args) {
		
		Scanner input=new Scanner(System.in);
		System.out.println("请输入一串数字：");
		String str=input.nextLine();
		System.out.println("请输入后移位置的m个数：");
		int m=input.nextInt();
		
		StringBuffer strr=new StringBuffer(str);
		
		strr.append(strr.substring(0, m));
		strr.delete(0,m);
		System.out.println(strr);
	}

}