我想创建一个URL缩短服务，你可以写一个长URL到输入字段和服务缩短URL为“http://www.example.org/abcdef”。
可以用包含a-z, a-z和0-9的6个字符的字符串代替"abcdef"。这样就有560 ~ 570亿个可能的字符串。
我的方法:
我有一个有三列的数据库表:
Id，整数，自动递增
long, string，用户输入的长URL
短，字符串，缩短的URL(或只有六个字符)
然后将长URL插入到表中。然后，我将为“id”选择自动递增的值，并构建它的散列。这个散列应该作为“short”插入。但是我应该构建什么样的哈希呢?像MD5这样的哈希算法会创建太长的字符串。我认为我不用这些算法。一个自建的算法也可以。
我的想法:
对于“http://www.google.de/”，我得到了自动增量id 239472。然后我执行以下步骤:
short = '';
if divisible by 2, add "a"+the result to short
if divisible by 3, add "b"+the result to short
... until I have divisors for a-z and A-Z.
可以不断重复，直到这个数不能被整除为止。你认为这是一个好方法吗?你有更好的主意吗?
由于对这个主题的持续兴趣，我已经发布了一个高效的GitHub解决方案，包括JavaScript、PHP、Python和Java的实现。如果你喜欢，添加你的解决方案:)
非常好的回答，我已经创建了一个bjf的Golang实现:
package bjf
import (
"math"
"strings"
"strconv"
)
const alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
func Encode(num string) string {
n, _ := strconv.ParseUint(num, 10, 64)
t := make([]byte, 0)
/* Special case */
if n == 0 {
return string(alphabet[0])
}
/* Map */
for n > 0 {
r := n % uint64(len(alphabet))
t = append(t, alphabet[r])
n = n / uint64(len(alphabet))
}
/* Reverse */
for i, j := 0, len(t) - 1; i < j; i, j = i + 1, j - 1 {
t[i], t[j] = t[j], t[i]
}
return string(t)
}
func Decode(token string) int {
r := int(0)
p := float64(len(token)) - 1
for i := 0; i < len(token); i++ {
r += strings.Index(alphabet, string(token[i])) * int(math.Pow(float64(len(alphabet)), p))
p--
}
return r
}
托管在github: https://github.com/xor-gate/go-bjf
alphabet = map(chr, range(97,123)+range(65,91)) + map(str,range(0,10))
def lookup(k, a=alphabet):
if type(k) == int:
return a[k]
elif type(k) == str:
return a.index(k)
def encode(i, a=alphabet):
'''Takes an integer and returns it in the given base with mappings for upper/lower case letters and numbers 0-9.'''
try:
i = int(i)
except Exception:
raise TypeError("Input must be an integer.")
def incode(i=i, p=1, a=a):
# Here to protect p.
if i <= 61:
return lookup(i)
else:
pval = pow(62,p)
nval = i/pval
remainder = i % pval
if nval <= 61:
return lookup(nval) + incode(i % pval)
else:
return incode(i, p+1)
return incode()
def decode(s, a=alphabet):
'''Takes a base 62 string in our alphabet and returns it in base10.'''
try:
s = str(s)
except Exception:
raise TypeError("Input must be a string.")
return sum([lookup(i) * pow(62,p) for p,i in enumerate(list(reversed(s)))])a
这是我的版本，给任何需要的人。
Scala中的实现:
class Encoder(alphabet: String) extends (Long => String) {
val Base = alphabet.size
override def apply(number: Long) = {
def encode(current: Long): List[Int] = {
if (current == 0) Nil
else (current % Base).toInt :: encode(current / Base)
}
encode(number).reverse
.map(current => alphabet.charAt(current)).mkString
}
}
class Decoder(alphabet: String) extends (String => Long) {
val Base = alphabet.size
override def apply(string: String) = {
def decode(current: Long, encodedPart: String): Long = {
if (encodedPart.size == 0) current
else decode(current * Base + alphabet.indexOf(encodedPart.head),encodedPart.tail)
}
decode(0,string)
}
}
使用Scala测试的测试示例:
import org.scalatest.{FlatSpec, Matchers}
class DecoderAndEncoderTest extends FlatSpec with Matchers {
val Alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
"A number with base 10" should "be correctly encoded into base 62 string" in {
val encoder = new Encoder(Alphabet)
encoder(127) should be ("cd")
encoder(543513414) should be ("KWGPy")
}
"A base 62 string" should "be correctly decoded into a number with base 10" in {
val decoder = new Decoder(Alphabet)
decoder("cd") should be (127)
decoder("KWGPy") should be (543513414)
}
}
