如果你像我一样想使用谷歌Guava而不是Apache Commons。你可以在Guava Strings类中使用repeat方法。

Strings.repeat("-", 60);

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")));

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!

所以你想避免循环?

现在你知道了:

public static String repeat(String s, int times) {
    if (times <= 0) return "";
    else return s + repeat(s, times-1);
}

(当然我知道这是丑陋和低效的，但它没有循环:-p)

你想要更简单更漂亮吗?使用jython:

s * 3

编辑:让我们优化一下:-D

public static String repeat(String s, int times) {
   if (times <= 0) return "";
   else if (times % 2 == 0) return repeat(s+s, times/2);
   else return s + repeat(s+s, times/2);
}

Edit2:我已经为4个主要的替代方案做了一个快速而粗略的基准测试，但我没有时间多次运行它来获得几个输入的平均值和绘制时间……如果有人想试试，下面是代码:

public class Repeat {
    public static void main(String[] args)  {
        int n = Integer.parseInt(args[0]);
        String s = args[1];
        int l = s.length();
        long start, end;

        start = System.currentTimeMillis();
        for (int i = 0; i < n; i++) {
            if(repeatLog2(s,i).length()!=i*l) throw new RuntimeException();
        }
        end = System.currentTimeMillis();
        System.out.println("RecLog2Concat: " + (end-start) + "ms");

        start = System.currentTimeMillis();
        for (int i = 0; i < n; i++) {
            if(repeatR(s,i).length()!=i*l) throw new RuntimeException();
        }               
        end = System.currentTimeMillis();
        System.out.println("RecLinConcat: " + (end-start) + "ms");

        start = System.currentTimeMillis();
        for (int i = 0; i < n; i++) {
            if(repeatIc(s,i).length()!=i*l) throw new RuntimeException();
        }
        end = System.currentTimeMillis();
        System.out.println("IterConcat: " + (end-start) + "ms");

        start = System.currentTimeMillis();
        for (int i = 0; i < n; i++) {
            if(repeatSb(s,i).length()!=i*l) throw new RuntimeException();
        }
        end = System.currentTimeMillis();
        System.out.println("IterStrB: " + (end-start) + "ms");
    }

    public static String repeatLog2(String s, int times) {
        if (times <= 0) {
            return "";
        }
        else if (times % 2 == 0) {
            return repeatLog2(s+s, times/2);
        }
        else {
           return s + repeatLog2(s+s, times/2);
        }
    }

    public static String repeatR(String s, int times) {
        if (times <= 0) {
            return "";
        }
        else {
            return s + repeatR(s, times-1);
        }
    }

    public static String repeatIc(String s, int times) {
        String tmp = "";
        for (int i = 0; i < times; i++) {
            tmp += s;
        }
        return tmp;
    }

    public static String repeatSb(String s, int n) {
        final StringBuilder sb = new StringBuilder();
        for(int i = 0; i < n; i++) {
            sb.append(s);
        }
        return sb.toString();
    }
}

它有两个参数，第一个是迭代的次数(每个函数运行的重复次数参数从1..n)，第二个是要重复的字符串。

到目前为止，快速检查不同输入的运行时间，得出的排名大致如下(从好到坏):

迭代StringBuilder追加(1x)。 递归连接log2调用(~3x)。 递归级联线性调用(~30x)。 迭代级联线性(~45x)。

我从来没有想到递归函数比for循环快:-o

玩得开心。

字符串:重复

". ".repeat(7)  // Seven period-with-space pairs: . . . . . . . 

Java 11中的新方法是String::repeat，它完全符合你的要求:

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

它的Javadoc说:

/**
 * Returns a string whose value is the concatenation of this
 * string repeated {@code count} times.
 * <p>
 * If this string is empty or count is zero then the empty
 * string is returned.
 *
 * @param count number of times to repeat
 *
 * @return A string composed of this string repeated
 * {@code count} times or the empty string if this
 * string is empty or count is zero
 *
 * @throws IllegalArgumentException if the {@code count} is
 * negative.
 *
 * @since 11
 */ 

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();
}