java根据生成随机数功能类编写程序

根据日生成随机数的功能类及方法编写程度，生成四位随机验证码的功能，Random类的nextInt(26）方法获取到不超过26的随机值；‘A’+n；当n不超过26时，为A-Z之间的字母值。Ps:小弟还是新手，希望大神帮帮我，谢谢啦！

private static String[] chars={"0","1","2","3","4","5","6","7","8","9","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"};
private static String getRandomChar(){
Random ran = new Random();
int index = ran.nextInt(chars.length);
return chars[index];
}
我意见这样生成验证码，代码可以复用~
你只要把数组中的0~9去掉就行了，验证码往往会去掉 0 ，1 ，o这些容易混淆的字符

   public static void main(string []args) throws exception{   linkedlist<integer> list = new linkedlist<integer>();   for(int i=0;i<10;i++){   list.add((int)(math.random()*integer.max_value));   }   while(list.size() > 0)   system.out.print(list.pop()+ " ");   }