java - 随机替换字符串中随机数的字符

Question

我是 Java 新手，我的代码需要一些帮助。

import java.util.Arrays;
import java.util.Scanner;
import java.util.Random;
public class asdf {
    public static void main(String[] args)
    {
        Random generator = new Random();
        int num1;
        String theWord;
        Scanner in = new Scanner(System.in);
        System.out.println("Enter a string to rearrange");
        theWord = in.nextLine();
        int length = theWord.length();
        Random times=new Random();
        System.out.println("Length of String:"+length);
        char[] chars=theWord.toCharArray();
        Arrays.sort(chars);
        int timez=times.nextInt();
        num1=generator.nextInt(length);
        char c=theWord.charAt(num1);
        String newWord=new String(chars);
        for(int i=0;i+=1;i<timez)
        {
            newWord=newWord.replace(c, '*');
        }
        System.out.println(newWord);
     }
}

所以我试图重新排列字符串并用“*”替换一些字符，但它不起作用，我不知道如何修复它。

Answer 1

import java.util.Arrays;
import java.util.Scanner;
import java.util.Random;
public class asdf {
public static void main(String[] args) {
    Random generator = new Random();
    int num1;
    String theWord;
    Scanner in = new Scanner(System.in);
    System.out.println("Enter a string to rearrange");
    theWord = in.nextLine();
    int length = theWord.length();
    Random times=new Random();
    System.out.println("Length of String:"+length);
    char[] chars=theWord.toCharArray();
    Arrays.sort(chars);
    int timez=times.nextInt();
    num1=generator.nextInt(length);
    char c=theWord.charAt(num1);
    String newWord=new String(chars);
    for(int i=0;i<timez;i++) {
        newWord=newWord.replace(c, '*');
    }
    System.out.println(newWord);
    }
 }

IDEONE 示例:http://ideone.com/PjfAor