Question

我有一个看起来像的字符串：

"Hello my is Joeseph. It is very nice to meet you. What a wonderful day it is!".

我想计算字符串中is的次数。

我怎样才能用Java做到这一点？

Answer 1

一种简单的方法是使用Apache StringUtils countMatches

StringUtils.countMatches("Hello my is Joeseph. It is very nice to meet you. What a wonderful day it is!", "is");

Answer 2

int index = input.indexOf("is");
int count = 0;
while (index != -1) {
    count++;
    input = input.substring(index + 1);
    index = input.indexOf("is");
}
System.out.println("No of *is* in the input is : " + count);

Answer 3

如果您更喜欢正则表达式，这是一个正则表达式解决方案：

String example = "Hello my is Joeseph. It is very nice to meet you. isWhat a wonderful day it is!";
Matcher m = Pattern.compile("\\bis\\b").matcher(example);

int matches = 0;
while(m.find())
    matches++;

System.out.println(matches);

在这种情况下，“isWhat”中的“is”被忽略，因为模式中的\ b边界匹配器

Answer 4

String haystack = "Hello my is Joeseph. It is very nice to meet you. What a wonderful day it is!";
haystack.toLowerCase();
String needle = "is";

int numNeedles = 0;

int pos = haystack.indexOf(needle);

    while(pos >= 0 ){

      pos = pos + 1;
      numNeedles = numNeedles + 1;

      pos = haystack.indexOf(needle,pos);

    }

 System.out.println("the num of " +needle+ "= " +numNeedles);

Answer 5

这考虑了“替换”的长度

String text = "a.b.c.d";
String replace = ".";
int count = (text.length()- (text.replaceAll(replace, "").length())) / replace.length();
System.out.println(count)

Answer 6

拆分每个“”（空白）并用循环检查输出的字符串[]

Answer 7

您可以找到code here。奇怪的是，它看起来像是罗比的那个。

如何计算序列在Java字符串中出现的次数？

7 个答案: