需求分析
从业务需求上来看,邀请码有以下几个强制性的要求:
- 不可重复
- 唯一确定
这两点要求首先就排除了 hash code 的可能,因为 hash code 是可以发生碰撞的。然后在强制性要求的基础之上,我们还有一些进一步的需求:
- 长度不能太长,6-10 位是合适的区间
- 不容易被推测出
- 资源消耗尽可能小
在这些需求的约束下,我们来看看相对来说比较通用的邀请码生成算法。
C#版本
public class InvitationCodeUtil { /// <summary> /// 随机字符串 /// </summary> private static readonly char[] CHARS = new char[] { 'F', 'L', 'G', 'W', '5', 'X', 'C', '3', '9', 'Z', 'M', '6', '7', 'Y', 'R', 'T', '2', 'H', 'S', '8', 'D', 'V', 'E', 'J', '4', 'K', 'Q', 'P', 'U', 'A', 'N', 'B' }; private const int CHARS_LENGTH = 32; /// <summary> /// 邀请码长度 /// </summary> private const int CODE_LENGTH = 8; /// <summary> /// 随机数据 /// </summary> private const long SLAT = 1234561L; /// <summary> /// PRIME1 与 CHARS 的长度 L互质,可保证 ( id * PRIME1) % L 在 [0,L)上均匀分布 /// </summary> private const int PRIME1 = 3; /// <summary> /// PRIME2 与 CODE_LENGTH 互质,可保证 ( index * PRIME2) % CODE_LENGTH 在 [0,CODE_LENGTH)上均匀分布 /// </summary> private const int PRIME2 = 11; /// <summary> /// 生成邀请码, 默认为 8位邀请码 /// </summary> /// <param name="id"> 唯一的id. 支持的最大值为: (32^7 - <seealso cref="SLAT"/>)/<seealso cref="PRIME1"/> = 11452834602 /// @return </param> public static string gen(long? id) { return gen(id, CODE_LENGTH); } /// <summary> /// 生成邀请码 /// </summary> /// <param name="id"> 唯一的id主键. 支持的最大值为: (32^7 - <seealso cref="SLAT"/>)/<seealso cref="PRIME1"/> </param> /// <returns> code </returns> public static string gen(long? id, int length) { // 补位 id = id * PRIME1 + SLAT; //将 id 转换成32进制的值 long[] b = new long[CODE_LENGTH]; // 32进制数 b[0] = id.Value; for (int i = 0; i < CODE_LENGTH - 1; i++) { b[i + 1] = b[i] / CHARS_LENGTH; // 按位扩散 b[i] = (b[i] + i * b[0]) % CHARS_LENGTH; } long tmp = 0; for (int i = 0; i < length - 2; i++) { tmp += b[i]; } b[length - 1] = tmp * PRIME1 % CHARS_LENGTH; // 进行混淆 long[] codeIndexArray = new long[CODE_LENGTH]; for (int i = 0; i < CODE_LENGTH; i++) { codeIndexArray[i] = b[i * PRIME2 % CODE_LENGTH]; } StringBuilder buffer = new StringBuilder(); //Arrays.stream(codeIndexArray).boxed().map(long?.intValue).map(t => CHARS[t]).forEach(buffer.append); foreach (var t in codeIndexArray) { buffer.Append(CHARS[t]); } return buffer.ToString(); } /// <summary> /// 将邀请码解密成原来的id /// </summary> /// <param name="code"> 邀请码 </param> /// <returns> id </returns> public static long? decode(string code) { if (code.Length != CODE_LENGTH) { return null; } // 将字符还原成对应数字 long[] a = new long[CODE_LENGTH]; for (int i = 0; i < CODE_LENGTH; i++) { char c = code[i]; int index = findIndex(c); if (index == -1) { // 异常字符串 return null; } a[i * PRIME2 % CODE_LENGTH] = index; } long[] b = new long[CODE_LENGTH]; for (int i = CODE_LENGTH - 2; i >= 0; i--) { b[i] = (a[i] - a[0] * i + CHARS_LENGTH * i) % CHARS_LENGTH; } long res = 0; for (int i = CODE_LENGTH - 2; i >= 0; i--) { res += b[i]; res *= (i > 0 ? CHARS_LENGTH : 1); } return (res - SLAT) / PRIME1; } /// <summary> /// 查找对应字符的index /// </summary> /// <param name="c"> 字符 </param> /// <returns> index </returns> public static int findIndex(char c) { for (int i = 0; i < CHARS_LENGTH; i++) { if (CHARS[i] == c) { return i; } } return -1; } }
Java版本:
public class InvitationCodeUtil { /** * 随机字符串 */ private static final char[] CHARS = new char[] {'F', 'L', 'G', 'W', '5', 'X', 'C', '3', '9', 'Z', 'M', '6', '7', 'Y', 'R', 'T', '2', 'H', 'S', '8', 'D', 'V', 'E', 'J', '4', 'K', 'Q', 'P', 'U', 'A', 'N', 'B'}; private final static int CHARS_LENGTH = 32; /** * 邀请码长度 */ private final static int CODE_LENGTH = 8; /** * 随机数据 */ private final static long SLAT = 1234561L; /** * PRIME1 与 CHARS 的长度 L互质,可保证 ( id * PRIME1) % L 在 [0,L)上均匀分布 */ private final static int PRIME1 = 3; /** * PRIME2 与 CODE_LENGTH 互质,可保证 ( index * PRIME2) % CODE_LENGTH 在 [0,CODE_LENGTH)上均匀分布 */ private final static int PRIME2 = 11; /** * 生成邀请码, 默认为 8位邀请码 * * @param id 唯一的id. 支持的最大值为: (32^7 - {@link #SLAT})/{@link #PRIME1} = 11452834602 * @return */ public static String gen(Long id) { return gen(id, CODE_LENGTH); } /** * 生成邀请码 * * @param id 唯一的id主键. 支持的最大值为: (32^7 - {@link #SLAT})/{@link #PRIME1} * @return code */ public static String gen(Long id, int length) { // 补位 id = id * PRIME1 + SLAT; //将 id 转换成32进制的值 long[] b = new long[CODE_LENGTH]; // 32进制数 b[0] = id; for (int i = 0; i < CODE_LENGTH - 1; i++) { b[i + 1] = b[i] / CHARS_LENGTH; // 按位扩散 b[i] = (b[i] + i * b[0]) % CHARS_LENGTH; } long tmp = 0; for (int i = 0; i < length - 2; i++) { tmp += b[i]; } b[length - 1] = tmp * PRIME1 % CHARS_LENGTH; // 进行混淆 long[] codeIndexArray = new long[CODE_LENGTH]; for (int i = 0; i < CODE_LENGTH; i++) { codeIndexArray[i] = b[i * PRIME2 % CODE_LENGTH]; } StringBuilder buffer = new StringBuilder(); Arrays.stream(codeIndexArray).boxed().map(Long::intValue).map(t -> CHARS[t]).forEach(buffer::append); return buffer.toString(); } /** * 将邀请码解密成原来的id * * @param code 邀请码 * @return id */ public static Long decode(String code) { if (code.length() != CODE_LENGTH) { return null; } // 将字符还原成对应数字 long[] a = new long[CODE_LENGTH]; for (int i = 0; i < CODE_LENGTH; i++) { char c = code.charAt(i); int index = findIndex(c); if (index == -1) { // 异常字符串 return null; } a[i * PRIME2 % CODE_LENGTH] = index; } long[] b = new long[CODE_LENGTH]; for (int i = CODE_LENGTH - 2; i >= 0; i--) { b[i] = (a[i] - a[0]*i + CHARS_LENGTH * i) % CHARS_LENGTH; } long res = 0; for (int i = CODE_LENGTH - 2; i >= 0; i--) { res += b[i]; res *= (i > 0 ? CHARS_LENGTH : 1); } return (res - SLAT) / PRIME1; } /** * 查找对应字符的index * * @param c 字符 * @return index */ public static int findIndex(char c) { for (int i = 0; i < CHARS_LENGTH; i++) { if (CHARS[i] == c) { return i; } } return -1; } }
下载 http://files.cnblogs.com/files/hongshao/JavatoCSharpConverter.Patched.rar
源文:https://www.jianshu.com/p/c250ccc5fe1e
趣谈唯一邀请码生成方法 https://my.oschina.net/bravozu/blog/1827254
简单的密码学生成唯一邀请码 https://huzb.me/2018/03/23/%E7%AE%80%E5%8D%95%E7%9A%84%E5%AF%86%E7%A0%81%E5%AD%A6%E7%94%9F%E6%88%90%E5%94%AF%E4%B8%80%E9%82%80%E8%AF%B7%E7%A0%81/