Question

我们有一个字符串输入，以下组合有效（例如sunday，*sunday*，sun*day*，*sun*day，su*nda*y）如果它只包含一个星号，那么这是一个错误的输入。

因此，鉴于上述输入，我如何检查字符串是否包含多个星号。

Answer 1

int asterisk1 = input.indexOf('*');
boolean hasTowAsterisks = asterisk1 != -1 && input.indexOf('*', asterisk1+1) != -1;

编辑：此解决方案假设要求“至少有两个星号”。

Answer 2

您可以将String.matches与正则表达式一起使用：

"^.*(?:\\*.*){2}$"

如果您想要完全两个星号：

"^[^*]*(?:\\*[^*]*){2}$"

虽然对于这个任务来说，迭代字符串并计算星号可能更简单。

Answer 3

至少有两种方式：

正则表达式

String regex = ".*\\*.*\\*.*";
boolean valid = input.matches(regex);

环

int asterisks = 0;
for (int i = 0; i < input.length(); i ++) {
    if (input.charAt(i) == '*') {
        asterisks++;
    }
}

Answer 4

这是一个适用于任何文字字符串的非正则表达式：

public static boolean containsNoneOrTwo(String haystack, String needle) {
    int index = haystack.indexOf(needle);
    return (index == -1) ||
        haystack.indexOf(needle, index+1) == haystack.lastIndexOf(needle);
}

基本上算法是：

containsNoneOrTwo(haystack, needle)
   = haystack contains no needle OR
        haystack's second needle is also its last

Answer 5

boolean hasTwoAsteriks(String str) {
int i;
if((i = str.indexOf("*")) != -1) {
  if ((i = str.indexOf("*", i+1)) != -1)
    return true;
return false;
}

Answer 6

为了完整性（尽管已经提供了几个好的答案，我喜欢Mark和Joachim的最佳答案），这里有两个基于String.split(regex)和String.split(regex, limit)的版本：

（编辑，错误修复:)

boolean containsAtLeastTwoAsterisks = ("_" + myString + "_").split("\\*", 3).length == 3;

boolean containsExactlyTwoAsterisks = ("_" + myString + "_").split("\\*").length == 3;

我根据我们的答案写了a little benchmark（我知道，基准测试并不重要，但它们很有趣，我知道可能是废话。）无论如何，以下是样本运行的结果：

*********************************************************************************

Testing strings with one or less asterisk

Processor: bhups
Finished. Duration: 40 ms, errors: 0

Processor: Bozho (loop version)
Finished. Duration: 33 ms, errors: 0

Processor: Bozho (regex version)
Finished. Duration: 806 ms, errors: 0

Processor: Joachim Sauer
Finished. Duration: 24 ms, errors: 0 <-- winner

Processor: Mark Byers
Finished. Duration: 1068 ms, errors: 0

Processor: seanizer
Finished. Duration: 408 ms, errors: 0

*********************************************************************************

Testing strings with exactly two asterisks

Processor: bhups
Finished. Duration: 14 ms, errors: 0 <-- winner

Processor: Bozho (loop version)
Finished. Duration: 21 ms, errors: 0

Processor: Bozho (regex version)
Finished. Duration: 693 ms, errors: 0

Processor: Joachim Sauer
Finished. Duration: 14 ms, errors: 0 <-- winner

Processor: Mark Byers
Finished. Duration: 491 ms, errors: 0

Processor: seanizer
Finished. Duration: 340 ms, errors: 0

*********************************************************************************

Testing strings with more than two asterisks (not all processors will be included)

Skipping processor bhups

Processor: Bozho (loop version)
Finished. Duration: 63 ms, errors: 0  <-- winner

Skipping processor Bozho (regex version)

Skipping processor Joachim Sauer

Processor: Mark Byers
Finished. Duration: 1555 ms, errors: 0

Processor: seanizer
Finished. Duration: 860 ms, errors: 0

似乎非正则表达式比正则表达式快得多。我猜这就是你所期望的。

编辑：修正了错误的赢家。抱歉，joachim

如何检查字符串是否包含两个星号字符？

6 个答案: