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问题描述
什么是洗牌一句话这是一个数组中的字母最简单的方法?我有一个数组一些话,我随机选择一个字,但我也想洗牌它的字母。
公共静态无效的主要(字串[] args){
的String [] =动物{狗,猫,恐龙};
随机随机=新的随机();
串词=动物[random.nextInt(animals.length)]; 的System.out.println(字);
//我只是想打乱单词的字母
}
我不应该使用该列表的事情。我想出了这样的事情,但这个code它打印随机字母不洗牌。也许我可以code类似,如果已经打印的信,不打印?
//一些随机的信
的for(int i = 0; I< word.length();我++){焦C =(word.charAt(random.nextInt(word.length())));
System.out.print(C); }
}
解决方案
真是没有必要收集和什么比接下来更多:
公共静态无效的主要(字串[] args){ //创建一个随机对象
随机R =新的随机(); 串词=动物; 的System.out.println(之前:+字);
字=争夺(R,字);
的System.out.println(后+字);
}公共静态字符串争抢(随机随机字符串inputString)
{
//将您的字符串转换成一个简单的字符数组:
所以char a [] = inputString.toCharArray(); //使用的争夺标准的费雪耶茨洗牌的字母,
的for(int i = 0; I<则为a.length-1;我++)
{
INT J = random.nextInt(则为a.length-1);
//交换信件
焦炭TEMP = a [i];一个由[i] = A [J]。一个研究[J] =温度;
} 返回新的String(一);
}
What is the easiest way to shuffle the letters of a word which is in an array? I have some words in an array and I choose randomly a word but I also want to shuffle the letters of it.
public static void main (String[] args ) {
String [] animals = { "Dog" , "Cat" , "Dino" } ;
Random random = new Random();
String word = animals [random.nextInt(animals.length)];
System.out.println ( word ) ;
//I Simply want to shuffle the letters of word
}
I am not supposed to use that List thing. I've come up with something like this but with this code it prints random letters it doesn't shuffle. Maybe I can code something like do not print if that letter already printed?
//GET RANDOM LETTER
for (int i = 0; i< word.length(); i++ ) {
char c = (word.charAt(random.nextInt(word.length())));
System.out.print(c); }
}
解决方案
Really no need for collection and anything more than what follows:
public static void main(String[] args) {
// Create a random object
Random r = new Random();
String word = "Animals";
System.out.println("Before: " + word );
word = scramble( r, word );
System.out.println("After : " + word );
}
public static String scramble( Random random, String inputString )
{
// Convert your string into a simple char array:
char a[] = inputString.toCharArray();
// Scramble the letters using the standard Fisher-Yates shuffle,
for( int i=0 ; i<a.length-1 ; i++ )
{
int j = random.nextInt(a.length-1);
// Swap letters
char temp = a[i]; a[i] = a[j]; a[j] = temp;
}
return new String( a );
}
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