要存儲在數組中的隨機字符發生器

Question

如何實例化我的數組，使其具有26個隨機字符？

import java.util.Random; 

public class Encryption { 

Random r = new Random(); 
char[] real = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'}; 
char[] decrypted = //?????? 

for(i=0; i < decrypted.length; i++) { 

} 

Answer 1

實例化大小爲26的數組;

char[] array = new char[26]; 

添加隨機字符

for(int i = 0; i < array.length; i++){ 
    array[i] = (char)(r.nextInt(26) +97); 
    System.out.print(array[i]); 
} 

（小寫字母）的範圍內可以修改此通過去除隨機部分，並使用i

for(int i = 0; i < array.length; i++){ 
    array[i] = (char)(i + 97); 
    System.out.print(array[i]); 
} 

Answer 2

你創建你的第一個數組應該檢查這個帖子。它看起來像你正在尋找的答案。 How to generate a random alpha-numeric string?

Answer 3

填寫您的例子：

import java.util.Random; 

public class Encryption { 

    Random r = new Random(); 
    char[] real = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'}; 
    char[] decrypted = new char[26]; 

    public static void main(String[] args){ 
    for(i=0; i < decrypted.length; i++) { 
     decrypted[i] = real[r.nextInt(26)]; 
    } 
    } 
}