如何在 java 中使用哈希集来确定字符串是否包含有效字符？
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How can I use hash sets in java to determine if a string contains valid characters?
提问by user2975289
I'm writing a lexical analyzer and have never used hash sets. I want to take a string and make sure it's legal. I think I understand how to build the hash set with valid characters but I'm not sure how to compare the string with teh hash set to ensure it contains valid characters. I can't find an example anywhere. Can someone point me to code that would do this?
我正在编写一个词法分析器并且从未使用过哈希集。我想拿一个字符串并确保它是合法的。我想我了解如何使用有效字符构建哈希集，但我不确定如何将字符串与哈希集进行比较以确保它包含有效字符。我在任何地方都找不到示例。有人可以指出我可以做到这一点的代码吗？
回答by wvdz
HashSet has the function contains() for this, since it implements the Collection interface.
HashSet 具有用于此的函数 contains()，因为它实现了 Collection 接口。
回答by dasblinkenlight
You cannot compare an entire string to a HashSet<Character>, but you can do it one character at a time:
您不能将整个字符串与 a 进行比较HashSet<Character>，但您可以一次做一个字符：
HashSet<Character> valid = new HashSet<Character>();
valid.add('a');
valid.add('d');
valid.add('f');
boolean allOk = true;
for (char c : "fad".toCharArray()) {
if (!valid.contains(c)) {
allOk = false;
break;
}
}
System.out.println(allOk);
However, this is not the most efficient way of doing it. A better approach would be to construct a regex with the characters that you need, and call match()on the string:
然而，这并不是最有效的方法。更好的方法是使用您需要的字符构造一个正则表达式，然后调用match()字符串：
// Let's say x, y, and z are the valid characters
String regex = "[xyz]*";
if (myString.matches(regex)) {
System.out.println("All characters in the string are in 'x', 'y', and 'z'");
}
回答by Stephen C
I think you are probably over-thinking this problem. (For instance, spending too much time thinking how to make the lexer "efficient" ...)
我想你可能把这个问题想得太多了。（例如，花太多时间思考如何使词法分析器“高效”......）
The conventional ways to test for valid / invalid characters in a lexer are:
在词法分析器中测试有效/无效字符的常规方法是：
use a big switch statement, or
perform a sequence of "character class" tests; e.g. using the result of
Character.getType(char)
使用大的 switch 语句，或者
执行一系列“字符类”测试；例如使用结果
Character.getType(char)
Or better still, use a lexer generator.
或者更好的是，使用词法分析器生成器。
Using a HashSet is neither more efficient or more readable than a switch. And the "character class" approach couldbe a lot more readable than both ... depending on your validation rules.
使用 HashSet 既不比switch. 并且“字符类”方法可能比两者都更具可读性……这取决于您的验证规则。
But if I haven't convinced you, see @blinkenlights' Answer :-)
但如果我还没有说服你，请看@blinkenlights 的回答 :-)
