字符串中子字符串的出现

Question

为什么以下算法不会停止？(str是我正在搜索的字符串,findStr是我想要查找的字符串)

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while (lastIndex != -1) {
    lastIndex = str.indexOf(findStr,lastIndex);

    if( lastIndex != -1)
        count++;

    lastIndex += findStr.length();
}

System.out.println(count);

Answer 1

如何使用Apache Commons Lang的StringUtils.countMatches？

String str = "helloslkhellodjladfjhello";
String findStr = "hello";

System.out.println(StringUtils.countMatches(str, findStr));

那输出:

3

Answer 2

你lastIndex += findStr.length();被放在括号外面,造成一个无限循环(当没有发现时,lastIndex总是如此findStr.length()).

这是固定版本:

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while (lastIndex != -1) {

    lastIndex = str.indexOf(findStr, lastIndex);

    if (lastIndex != -1) {
        count++;
        lastIndex += findStr.length();
    }
}
System.out.println(count);

Answer 3

更短的版本.;)

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
System.out.println(str.split(findStr, -1).length-1);

Answer 4

最后一行是造成问题.lastIndex永远不会在-1,所以会有一个无限循环.这可以通过将最后一行代码移动到if块来修复.

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while(lastIndex != -1){

    lastIndex = str.indexOf(findStr,lastIndex);

    if(lastIndex != -1){
        count ++;
        lastIndex += findStr.length();
    }
}
System.out.println(count);

Answer 5

你真的必须自己处理匹配吗？特别是如果你需要的只是出现次数,正则表达式更整洁:

String str = "helloslkhellodjladfjhello";
Pattern p = Pattern.compile("hello");
Matcher m = p.matcher(str);
int count = 0;
while (m.find()){
    count +=1;
}
System.out.println(count);     

Answer 6

在这里,它包含在一个漂亮且可重用的方法中:

public static int count(String text, String find) {
        int index = 0, count = 0, length = find.length();
        while( (index = text.indexOf(find, index)) != -1 ) {                
                index += length; count++;
        }
        return count;
}

Answer 7

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while((lastIndex = str.indexOf(findStr, lastIndex)) != -1) {
     count++;
     lastIndex += findStr.length() - 1;
}
System.out.println(count);

循环结束时计数为3; 希望能帮助到你

Answer 8

我很惊讶没有人提到这一支班轮。它简单，简洁，并且比str.split(target, -1).length-1

public static int count(String str, String target) {
    return (str.length() - str.replace(target, "").length()) / target.length();
}

Answer 9

许多给定的答案在以下一个或多个方面失败:

• 任意长度的图案
• 重叠匹配(例如"23232"中的"232"或"aaa"中的"aa")
• 正则表达式元字符

这是我写的:

static int countMatches(Pattern pattern, String string)
{
    Matcher matcher = pattern.matcher(string);

    int count = 0;
    int pos = 0;
    while (matcher.find(pos))
    {
        count++;
        pos = matcher.start() + 1;
    }

    return count;
}

示例电话:

Pattern pattern = Pattern.compile("232");
int count = countMatches(pattern, "23232"); // Returns 2

如果你想要一个非正则表达式搜索,只需使用LITERAL标志适当地编译你的模式:

Pattern pattern = Pattern.compile("1+1", Pattern.LITERAL);
int count = countMatches(pattern, "1+1+1"); // Returns 2

Answer 10

您可以使用内置库函数的出现次数：

import org.springframework.util.StringUtils;
StringUtils.countOccurrencesOf(result, "R-")

Answer 11

public int countOfOccurrences(String str, String subStr) {
  return (str.length() - str.replaceAll(Pattern.quote(subStr), "").length()) / subStr.length();
}