你可以通过保留填充数据来减少每次调用的开销，而不是每次都重新构建:

public class RightPadder {

    private int length;
    private String padding;

    public RightPadder(int length, String pad) {
        this.length = length;
        StringBuilder sb = new StringBuilder(pad);
        while (sb.length() < length) {
            sb.append(sb);
        }
        padding = sb.toString();
   }

    public String pad(String s) {
        return (s.length() < length ? s + padding : s).substring(0, length);
    }

}

作为一种替代方法，您可以将结果长度作为pad(…)方法的参数。在这种情况下，在该方法中而不是在构造函数中调整隐藏填充。

(提示:为了获得额外的学分，让它是线程安全的!: -)

填充到10个字符:

String.format("%10s", "foo").replace(' ', '*');
String.format("%-10s", "bar").replace(' ', '*');
String.format("%10s", "longer than 10 chars").replace(' ', '*');

输出:

  *******foo
  bar*******
  longer*than*10*chars

密码字符显示“*”:

String password = "secret123";
String padded = String.format("%"+password.length()+"s", "").replace(' ', '*');

输出与密码字符串长度相同:

  secret123
  *********

这是一个高效的实用工具类，用于Java中的左填充，右填充，中心填充和零填充字符串。

package com.example;

/**
 * Utility class for left pad, right pad, center pad and zero fill.
 */
public final class StringPadding {

    public static String left(String string, int length, char fill) {

        if (string.length() < length) {

            char[] chars = string.toCharArray();
            char[] output = new char[length];

            int delta = length - chars.length;

            for (int i = 0; i < length; i++) {
                if (i < delta) {
                    output[i] = fill;
                } else {
                    output[i] = chars[i - delta];
                }
            }

            return new String(output);
        }

        return string;
    }

    public static String right(String string, int length, char fill) {

        if (string.length() < length) {

            char[] chars = string.toCharArray();
            char[] output = new char[length];

            for (int i = 0; i < length; i++) {
                if (i < chars.length) {
                    output[i] = chars[i];
                } else {
                    output[i] = fill;
                }
            }

            return new String(output);
        }

        return string;
    }

    public static String center(String string, int length, char fill) {

        if (string.length() < length) {

            char[] chars = string.toCharArray();

            int delta = length - chars.length;
            int a = (delta % 2 == 0) ? delta / 2 : delta / 2 + 1;
            int b = a + chars.length;

            char[] output = new char[length];
            for (int i = 0; i < length; i++) {
                if (i < a) {
                    output[i] = fill;
                } else if (i < b) {
                    output[i] = chars[i - a];
                } else {
                    output[i] = fill;
                }
            }

            return new String(output);
        }

        return string;
    }

    public static String zerofill(String string, int length) {
        return left(string, length, '0');
    }

    private StringPadding() {
    }

    /**
     * For tests!
     */
    public static void main(String[] args) {

        String string = "123";
        char blank = ' ';

        System.out.println("left pad:    [" + StringPadding.left(string, 10, blank) + "]");
        System.out.println("right pad:   [" + StringPadding.right(string, 10, blank) + "]");
        System.out.println("center pad:  [" + StringPadding.center(string, 10, blank) + "]");
        System.out.println("zero fill:   [" + StringPadding.zerofill(string, 10) + "]");
    }
}

输出如下:

left pad:    [       123]
right pad:   [123       ]
center pad:  [    123   ]
zero fill:   [0000000123]

另一种利用递归的解决方案。

这与所有JDK版本兼容，不需要外部库:

private static String addPadding(final String str, final int desiredLength, final String padBy) {
    String result = str;
    if (str.length() >= desiredLength) {
        return result;
    } else {
        result += padBy;
        return addPadding(result, desiredLength, padBy);
    }
}

注意:这个解决方案将附加填充，与一个小调整，你可以前缀填充值。

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

概括一下Eko的答案(Java 11+):

public class StringUtils {
    public static String padLeft(String s, char fill, int padSize) {
        if (padSize < 0) {
            var err = "padSize must be >= 0 (was " + padSize + ")";
            throw new java.lang.IllegalArgumentException(err);
        }

        int repeats = Math.max(0, padSize - s.length());
        return Character.toString(fill).repeat(repeats) + s;
    }

    public static String padRight(String s, char fill, int padSize) {
        if (padSize < 0) {
            var err = "padSize must be >= 0 (was " + padSize + ")";
            throw new java.lang.IllegalArgumentException(err);
        }

        int repeats = Math.max(0, padSize - s.length());
        return s + Character.toString(fill).repeat(repeats);
    }

    public static void main(String[] args) {
        System.out.println(padLeft("", 'x', 5)); // => xxxxx
        System.out.println(padLeft("1", 'x', 5)); // => xxxx1
        System.out.println(padLeft("12", 'x', 5)); // => xxx12
        System.out.println(padLeft("123", 'x', 5)); // => xx123
        System.out.println(padLeft("1234", 'x', 5)); // => x1234
        System.out.println(padLeft("12345", 'x', 5)); // => 12345
        System.out.println(padLeft("123456", 'x', 5)); // => 123456

        System.out.println(padRight("", 'x', 5)); // => xxxxx
        System.out.println(padRight("1", 'x', 5)); // => 1xxxx
        System.out.println(padRight("12", 'x', 5)); // => 12xxx
        System.out.println(padRight("123", 'x', 5)); // => 123xx
        System.out.println(padRight("1234", 'x', 5)); // => 1234x
        System.out.println(padRight("12345", 'x', 5)); // => 12345
        System.out.println(padRight("123456", 'x', 5)); // => 123456

        System.out.println(padRight("1", 'x', -1)); // => throws
    }
}