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
    }
}

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
    }
}

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。

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!: -)

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*

s =字符串填充 n =期望长度 c = Char到pad

private String padStart( String s, int n, char c ){
    if( n <= 0 ) return s;
    else return String.format("%" + n + "s", s).replace(' ', c);
}