两种JAVA实现短网址服务算法

static final char[] DIGITS = 
{ '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f', 'g',
'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x',
'y', 'z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O',
'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z' };

public String shorten(String longUrl, int urlLength) {
  if (urlLength < 0 || urlLength > 6) {
   throw new IllegalArgumentException("the length of url must be between 0 and 6");
  }
  String md5Hex = DigestUtils.md5Hex(longUrl);
  // 6 digit binary can indicate 62 letter & number from 0-9a-zA-Z
  int binaryLength = urlLength * 6;
  long binaryLengthFixer = Long.valueOf(StringUtils.repeat("1", binaryLength), BINARY);
  for (int i = 0; i < 4; i++) {
   String subString = StringUtils.substring(md5Hex, i * 8, (i + 1) * 8);
   subString = Long.toBinaryString(Long.valueOf(subString, 16) & binaryLengthFixer);
   subString = StringUtils.leftPad(subString, binaryLength, "0");
   StringBuilder sbBuilder = new StringBuilder();
   for (int j = 0; j < urlLength; j++) {
    String subString2 = StringUtils.substring(subString, j * 6, (j + 1) * 6);
    int charIndex = Integer.valueOf(subString2, BINARY) & NUMBER_61;
    sbBuilder.append(DIGITS[charIndex]);
   }
   String shortUrl = sbBuilder.toString();
   if (lookupLong(shortUrl) != null) {
    continue;
   } else {
    return shortUrl;
   }
  }
  // if all 4 possibilities are already exists
  return null;
 }

private AtomicLong sequence = new AtomicLong(0);

 @Override
 protected String shorten(String longUrl) {
  long myseq = sequence.incrementAndGet();
  String shortUrl = to62RadixString(myseq);
  return shortUrl;
 }

 private String to62RadixString(long seq) {
  StringBuilder sBuilder = new StringBuilder();
  while (true) {
   int remainder = (int) (seq % 62);
   sBuilder.append(DIGITS[remainder]);
   seq = seq / 62;
   if (seq == 0) {
    break;
   }
  }
  return sBuilder.toString();
 }