下面是另一种向右填充的方法:

// put the number of spaces, or any character you like, in your paddedString

String paddedString = "--------------------";

String myStringToBePadded = "I like donuts";

myStringToBePadded = myStringToBePadded + paddedString.substring(myStringToBePadded.length());

//result:
myStringToBePadded = "I like donuts-------";

不管怎样，我一直在寻找一些可以填充的东西，然后我决定自己编写代码。它非常简洁，你可以很容易地从中推导出padLeft和padRight

    /**
     * Pads around a string, both left and right using pad as the template, aligning to the right or left as indicated.
     * @param a the string to pad on both left and right
     * @param pad the template to pad with, it can be of any size
     * @param width the fixed width to output
     * @param alignRight if true, when the input string is of odd length, adds an extra pad char to the left, so values are right aligned
     *                   otherwise add an extra pad char to the right. When the input is of even length no extra chars will be inserted
     * @return the input param a padded around.
     */
    public static String padAround(String a, String pad, int width, boolean alignRight) {
        if (pad.length() == 0)
            throw new IllegalArgumentException("Pad cannot be an empty string!");
        int delta = width - a.length();
        if (delta < 1)
            return a;
        int half = delta / 2;
        int remainder = delta % 2;
        String padding = pad.repeat(((half+remainder)/pad.length()+1)); // repeating the padding to occupy all possible space
        StringBuilder sb = new StringBuilder(width);
//        sb.append( padding.substring(0,half + (alignRight ? 0 : remainder)));
        sb.append(padding, 0, half + (alignRight ? 0 : remainder));
        sb.append(a);
//        sb.append( padding.substring(0,half + (alignRight ? remainder : 0)));
        sb.append(padding, 0, half + (alignRight ? remainder : 0));

        return sb.toString();
    }

虽然它应该是相当快的，它可能会受益于使用一些韵母在这里和那里。

所有的字符串操作通常都需要非常高效——特别是当你处理大数据集的时候。我想要的东西是快速和灵活的，类似于你将在plsql pad命令。另外，我不想为一件小事包含一个巨大的库。考虑到这些因素，这些解决方案都不令人满意。这是我提出的解决方案，有最好的基准测试结果，如果有人可以改进它，请添加您的评论。

public static char[] lpad(char[] pStringChar, int pTotalLength, char pPad) {
    if (pStringChar.length < pTotalLength) {
        char[] retChar = new char[pTotalLength];
        int padIdx = pTotalLength - pStringChar.length;
        Arrays.fill(retChar, 0, padIdx, pPad);
        System.arraycopy(pStringChar, 0, retChar, padIdx, pStringChar.length);
        return retChar;
    } else {
        return pStringChar;
    }
}

注意它是用String. tochararray()调用的，结果可以用new String((char[])result)转换为String。这样做的原因是，如果你应用多个操作，你可以在char[]上完成它们，而不需要在格式之间进行转换——在幕后，字符串存储为char[]。如果这些操作包含在String类本身中，那么效率将提高一倍——速度和内存方面。

在Dzone上找到的

用零填充:

String.format("|%020d|", 93); // prints: |00000000000000000093|

public static String LPad(String str, Integer length, char car) {
  return (str + String.format("%" + length + "s", "").replace(" ", String.valueOf(car))).substring(0, length);
}

public static String RPad(String str, Integer length, char car) {
  return (String.format("%" + length + "s", "").replace(" ", String.valueOf(car)) + str).substring(str.length(), length + str.length());
}

LPad("Hi", 10, 'R') //gives "RRRRRRRRHi"
RPad("Hi", 10, 'R') //gives "HiRRRRRRRR"
RPad("Hi", 10, ' ') //gives "Hi        "
RPad("Hi", 1, ' ')  //gives "H"
//etc...

为此编写自己的函数可能比其他答案更简单。

private static String padRight(String str, String padChar, int n) {

    String paddedString = str;

    while (paddedString.length() < n) {

        paddedString = padChar + str;

    }

    return paddedString;

}

private static String padLeft(String str, String padChar, int n) {

    String paddedString = str;

    while (paddedString.length() < n) {

        paddedString += padChar;

    }

    return paddedString;

}