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。

使用该函数。

private String leftPadding(String word, int length, char ch) {
   return (length > word.length()) ? leftPadding(ch + word, length, ch) : word;
}

如何使用?

leftPadding(month, 2, '0');

输出: 01 02 03 04 ..11日12

你可以使用内置的StringBuilder append()和insert()方法， 对于可变字符串长度的填充:

AbstractStringBuilder append(CharSequence s, int start, int end) ;

例如:

private static final String  MAX_STRING = "                    "; //20 spaces

    Set<StringBuilder> set= new HashSet<StringBuilder>();
    set.add(new StringBuilder("12345678"));
    set.add(new StringBuilder("123456789"));
    set.add(new StringBuilder("1234567811"));
    set.add(new StringBuilder("12345678123"));
    set.add(new StringBuilder("1234567812234"));
    set.add(new StringBuilder("1234567812222"));
    set.add(new StringBuilder("12345678122334"));

    for(StringBuilder padMe: set)
        padMe.append(MAX_STRING, padMe.length(), MAX_STRING.length());

我知道这个线程有点老了，最初的问题是为了一个简单的解决方案，但如果它应该是真的很快，你应该使用字符数组。

public static String pad(String str, int size, char padChar)
{
    if (str.length() < size)
    {
        char[] temp = new char[size];
        int i = 0;

        while (i < str.length())
        {
            temp[i] = str.charAt(i);
            i++;
        }

        while (i < size)
        {
            temp[i] = padChar;
            i++;
        }

        str = new String(temp);
    }

    return str;
}

格式化程序解决方案不是最佳的。仅仅构建格式字符串就会创建2个新字符串。

Apache的解决方案可以通过用目标大小初始化sb来改进，从而替换下面的内容

StringBuffer padded = new StringBuffer(str); 

与

StringBuffer padded = new StringBuffer(pad); 
padded.append(value);

会阻止某人内部缓冲的增长。

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;

}

从Java 1.5开始，string. format()可以用于左/右填充给定的字符串。

public static String padRight(String s, int n) {
     return String.format("%-" + n + "s", s);  
}

public static String padLeft(String s, int n) {
    return String.format("%" + n + "s", s);  
}

...

public static void main(String args[]) throws Exception {
 System.out.println(padRight("Howto", 20) + "*");
 System.out.println(padLeft("Howto", 20) + "*");
}

输出为:

Howto               *
               Howto*