如何检查一个字符串是否包含两个星号字符？

Question

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

因此，鉴于上述输入，如何检查字符串是否包含多个星号。

Answer 1

你可以使用String.matches使用正则表达式：

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

如果你想正是两个星号：

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

虽然这项任务可能是简单的只是遍历字符串和计数星号。

Answer 2

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 3

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

编辑：此解决方案假定要求是“至少有两个星号”。

Answer 4

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

Answer 5

这里是与任何文字字符串工作的非正则表达式的选择：

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

本质的算法是：

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

Answer 6

为了完整（虽然有几个很好的答案已经被提供，我喜欢Mark's和Joachim的最好），这里有两个版本，基于String.split(regex)和String.split(regex, limit)：

（编辑，bug修复:)

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

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

---

我根据我们的答案（我知道，基准意义已经不大，但它们很有趣，而我可能是废话，我知道写a little benchmark ）无论如何，这里是一个样本运行的结果：

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

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很多。这就是你期望的，我猜。

编辑：修正了错误的赢家。对不起，约阿希姆