在番石榴中，这很简单:

Strings.padStart("string", 10, ' ');
Strings.padEnd("string", 10, ' ');

一个简单的解决方案是:

package nl;
public class Padder {
    public static void main(String[] args) {
        String s = "123" ;
        System.out.println("#"+("     " + s).substring(s.length())+"#");
    }
}

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...

从Java 11开始，string. repeat(int)可以用来左右填充给定的字符串。

System.out.println("*".repeat(5)+"apple");
System.out.println("apple"+"*".repeat(5));

输出:

*****apple
apple*****

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
    }
}

这里有一个并行版本的你有很长的字符串:-)

int width = 100;
String s = "129018";

CharSequence padded = IntStream.range(0,width)
            .parallel()
            .map(i->i-(width-s.length()))
            .map(i->i<0 ? '0' :s.charAt(i))
            .collect(StringBuilder::new, (sb,c)-> sb.append((char)c), (sb1,sb2)->sb1.append(sb2));