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
    }
}

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!: -)

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]

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());

Java联机程序，没有花哨的库。

// 6 characters padding example
String pad = "******";
// testcases for 0, 4, 8 characters
String input = "" | "abcd" | "abcdefgh"

左Pad，不要限制

result = pad.substring(Math.min(input.length(),pad.length())) + input;
results: "******" | "**abcd" | "abcdefgh"

右移，不要限制

result = input + pad.substring(Math.min(input.length(),pad.length()));
results: "******" | "abcd**" | "abcdefgh"

左衬垫，限制衬垫长度

result = (pad + input).substring(input.length(), input.length() + pad.length());
results: "******" | "**abcd" | "cdefgh"

右垫，限制垫的长度

result = (input + pad).substring(0, pad.length());
results: "******" | "abcd**" | "abcdef"

@ck和@Marlon Tarak的答案是唯一使用char[]的答案，对于每秒有几个填充方法调用的应用程序来说，这是最好的方法。然而，它们没有利用任何数组操作优化，而且对我来说有点覆盖;这完全不需要循环。

public static String pad(String source, char fill, int length, boolean right){
    if(source.length() > length) return source;
    char[] out = new char[length];
    if(right){
        System.arraycopy(source.toCharArray(), 0, out, 0, source.length());
        Arrays.fill(out, source.length(), length, fill);
    }else{
        int sourceOffset = length - source.length();
        System.arraycopy(source.toCharArray(), 0, out, sourceOffset, source.length());
        Arrays.fill(out, 0, sourceOffset, fill);
    }
    return new String(out);
}

简单测试方法:

public static void main(String... args){
    System.out.println("012345678901234567890123456789");
    System.out.println(pad("cats", ' ', 30, true));
    System.out.println(pad("cats", ' ', 30, false));
    System.out.println(pad("cats", ' ', 20, false));
    System.out.println(pad("cats", '$', 30, true));
    System.out.println(pad("too long for your own good, buddy", '#', 30, true));
}

输出:

012345678901234567890123456789
cats                          
                          cats
                cats
cats$$$$$$$$$$$$$$$$$$$$$$$$$$
too long for your own good, buddy