我正在寻找一个简单的公共方法或操作符,允许我重复一些字符串n次。我知道我可以使用for循环来写这个,但我希望在必要时避免for循环,并且应该在某个地方存在一个简单的直接方法。

String str = "abc";
String repeated = str.repeat(3);

repeated.equals("abcabcabc");

相关:

重复字符串javascript 通过重复给定次数的另一个字符串创建NSString

编辑

当它们不是完全必要的时候,我尽量避免使用for循环,因为:

They add to the number of lines of code even if they are tucked away in another function. Someone reading my code has to figure out what I am doing in that for loop. Even if it is commented and has meaningful variables names, they still have to make sure it is not doing anything "clever". Programmers love to put clever things in for loops, even if I write it to "only do what it is intended to do", that does not preclude someone coming along and adding some additional clever "fix". They are very often easy to get wrong. For loops involving indexes tend to generate off by one bugs. For loops often reuse the same variables, increasing the chance of really hard to find scoping bugs. For loops increase the number of places a bug hunter has to look.


当前回答

Java 8的字符串。join结合Collections.nCopies提供了一种简洁的方法:

// say hello 100 times
System.out.println(String.join("", Collections.nCopies(100, "hello")));

其他回答

简单的循环

public static String repeat(String string, int times) {
    StringBuilder out = new StringBuilder();
    while (times-- > 0) {
        out.append(string);
    }
    return out.toString();
}

我很喜欢这个问题。有很多知识和风格。所以我不能没有展示我的摇滚就离开;)

{
    String string = repeat("1234567890", 4);
    System.out.println(string);
    System.out.println("=======");
    repeatWithoutCopySample(string, 100000);
    System.out.println(string);// This take time, try it without printing
    System.out.println(string.length());
}

/**
 * The core of the task.
 */
@SuppressWarnings("AssignmentToMethodParameter")
public static char[] repeat(char[] sample, int times) {
    char[] r = new char[sample.length * times];
    while (--times > -1) {
        System.arraycopy(sample, 0, r, times * sample.length, sample.length);
    }
    return r;
}

/**
 * Java classic style.
 */
public static String repeat(String sample, int times) {
    return new String(repeat(sample.toCharArray(), times));
}

/**
 * Java extreme memory style.
 */
@SuppressWarnings("UseSpecificCatch")
public static void repeatWithoutCopySample(String sample, int times) {
    try {
        Field valueStringField = String.class.getDeclaredField("value");
        valueStringField.setAccessible(true);
        valueStringField.set(sample, repeat((char[]) valueStringField.get(sample), times));
    } catch (Exception ex) {
        throw new RuntimeException(ex);
    }
}

你喜欢吗?

public static String repeat(String str, int times) {
    int length = str.length();
    int size = length * times;
    char[] c = new char[size];
    for (int i = 0; i < size; i++) {
        c[i] = str.charAt(i % length);
    }
    return new String(c);
}
public static String rep(int a,String k)

       {
           if(a<=0)
                return "";
           else 
           {a--;
               return k+rep(a,k);
       }

你可以使用这种递归方法实现你想要的目标。

使用递归,你可以做以下事情(使用三元运算符,最多一行):

public static final String repeat(String string, long number) {
    return number == 1 ? string : (number % 2 == 0 ? repeat(string + string, number / 2) : string + repeat(string + string, (number - 1) / 2));
}

我知道,它很丑,而且可能效率不高,但它就是一行!