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();
}

基于fortran的答案，这是一个使用StringBuilder的递归版本:

public static void repeat(StringBuilder stringBuilder, String s, int times) {
    if (times > 0) {
        repeat(stringBuilder.append(s), s, times - 1);
    }
}

public static String repeat(String s, int times) {
    StringBuilder stringBuilder = new StringBuilder(s.length() * times);
    repeat(stringBuilder, s, times);
    return stringBuilder.toString();
}

尽管您不希望使用循环，但我认为您应该使用循环。

String repeatString(String s, int repetitions)
{
    if(repetitions < 0) throw SomeException();

    else if(s == null) return null;

    StringBuilder stringBuilder = new StringBuilder(s.length() * repetitions);

    for(int i = 0; i < repetitions; i++)
        stringBuilder.append(s);

    return stringBuilder.toString();
}

你不使用for循环的理由并不充分。针对你的批评:

Whatever solution you use will almost certainly be longer than this. Using a pre-built function only tucks it under more covers. Someone reading your code will have to figure out what you're doing in that non-for-loop. Given that a for-loop is the idiomatic way to do this, it would be much easier to figure out if you did it with a for loop. Yes someone might add something clever, but by avoiding a for loop you are doing something clever. That's like shooting yourself in the foot intentionally to avoid shooting yourself in the foot by accident. Off-by-one errors are also mind-numbingly easy to catch with a single test. Given that you should be testing your code, an off-by-one error should be easy to fix and catch. And it's worth noting: the code above does not contain an off-by-one error. For loops are equally easy to get right. So don't reuse variables. That's not the for-loop's fault. Again, so does whatever solution you use. And as I noted before; a bug hunter will probably be expecting you to do this with a for loop, so they'll have an easier time finding it if you use a for loop.

这是最新的Stringutils.java

    public static String repeat(String str, int repeat) {
    // Performance tuned for 2.0 (JDK1.4)

    if (str == null) {
        return null;
    }
    if (repeat <= 0) {
        return EMPTY;
    }
    int inputLength = str.length();
    if (repeat == 1 || inputLength == 0) {
        return str;
    }
    if (inputLength == 1 && repeat <= PAD_LIMIT) {
        return repeat(str.charAt(0), repeat);
    }

    int outputLength = inputLength * repeat;
    switch (inputLength) {
        case 1 :
            return repeat(str.charAt(0), repeat);
        case 2 :
            char ch0 = str.charAt(0);
            char ch1 = str.charAt(1);
            char[] output2 = new char[outputLength];
            for (int i = repeat * 2 - 2; i >= 0; i--, i--) {
                output2[i] = ch0;
                output2[i + 1] = ch1;
            }
            return new String(output2);
        default :
            StringBuilder buf = new StringBuilder(outputLength);
            for (int i = 0; i < repeat; i++) {
                buf.append(str);
            }
            return buf.toString();
    }
    }

它甚至不需要这么大，可以做成这样，可以复制和粘贴 到项目中的实用程序类中。

    public static String repeat(String str, int num) {
    int len = num * str.length();
    StringBuilder sb = new StringBuilder(len);
    for (int i = 0; i < times; i++) {
        sb.append(str);
    }
    return sb.toString();
    }

所以e5，我认为最好的方法是简单地使用上面提到的代码，或者这里的任何答案。但是如果它是一个小项目，Commons lang就太大了

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")));

Commons only StringUtils.repeat（）

用法:

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

repeated.equals("abcabcabc");