比较 Java 中的两个十六进制字符串？
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Compare two hex strings in Java?
提问by tree-hacker
I am implementing a simple DHT using the Chord protocol in Java. The details are not important but the thing I'm stuck on is I need to hash strings and then see if one hashed string is "less than" another.
我正在使用 Java 中的 Chord 协议实现一个简单的 DHT。细节并不重要，但我坚持的事情是我需要散列字符串，然后查看一个散列字符串是否“小于”另一个。
I have some code to compute hashes using SHA1 which returns a 40 digit long hex string (of type String in Java) such as:
我有一些代码可以使用 SHA1 计算哈希，它返回一个 40 位长的十六进制字符串（Java 中的 String 类型），例如：
69342c5c39e5ae5f0077aecc32c0f81811fb8193
However I need to be able to compare two of these so to tell, for example that:
但是，我需要能够比较其中的两个，以便告诉，例如：
0000000000000000000000000000000000000000
is less than:
小于：
FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
This is the complete range of values as the 40 digit string is actually representing 40 hex numbers in the range 0123456789ABCDEF
这是完整的值范围，因为 40 位字符串实际上代表 0123456789ABCDEF 范围内的 40 个十六进制数字
Does anyone know how to do this?
有谁知道如何做到这一点？
Thanks in advance.
提前致谢。
回答by James Cronen
The values 0..9and A..Fare in hex-digit order in the ASCII character set, so
值0..9和A..F在 ASCII 字符集中按十六进制数字顺序排列，因此
string1.compareTo(string2)
should do the trick. Unless I'm missing something.
应该做的伎俩。除非我遗漏了什么。
回答by Adam
BigInteger one = new BigInteger("FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",16);
BigInteger two = new BigInteger("0000000000000000000000000000000000000000",16);
System.out.println(one.compareTo(two));
System.out.println(two.compareTo(one));
Output:
1
-1
输出：
1
-1
1 indicates greater than -1 indicates less than 0 would indicate equal values
1 表示大于 -1 表示小于 0 表示相等
回答by Adrian Pronk
Since hex characters are in ascending ascii order (as @Tenner indicated), you can directly compare the strings:
由于十六进制字符按 ascii 升序排列（如@Tenner 所示），您可以直接比较字符串：
String hash1 = ...;
String hash2 = ...;
int comparisonResult = hash1.compareTo(hash2);
if (comparisonResult < 0) {
// hash1 is less
}
else if (comparisonResult > 0) {
// hash1 is greater
}
else {
// comparisonResult == 0: hash1 compares equal to hash2
}
回答by vitaut
Since the strings are fixed length and '0' < '1' < ... < 'A' < ... < 'Z' you can use compareTo. If you use mixed case hex digits use compareToIgnoreCase.
由于字符串是固定长度的，并且 '0' < '1' < ... < 'A' < ... < 'Z' 您可以使用compareTo. 如果您使用大小写混合的十六进制数字，请使用compareToIgnoreCase.
