Occurrences of substring in a string

Question

Why is the following algorithm not halting for me? (str is the string I am searching in, findStr is the string I am trying to find)

String str = \"helloslkhe         


        

Answer 1

This below method show how many time substring repeat on ur whole string. Hope use full to you:-

    String searchPattern="aaa"; // search string
    String str="aaaaaababaaaaaa"; // whole string
    int searchLength = searchPattern.length(); 
    int totalLength = str.length(); 
    int k = 0;
    for (int i = 0; i < totalLength - searchLength + 1; i++) {
        String subStr = str.substring(i, searchLength + i);
        if (subStr.equals(searchPattern)) {
           k++;
        }

    }

Answer 2

here is the other solution without using regexp/patterns/matchers or even not using StringUtils.

String str = "helloslkhellodjladfjhelloarunkumarhelloasdhelloaruhelloasrhello";
        String findStr = "hello";
        int count =0;
        int findStrLength = findStr.length();
        for(int i=0;i<str.length();i++){
            if(findStr.startsWith(Character.toString(str.charAt(i)))){
                if(str.substring(i).length() >= findStrLength){
                    if(str.substring(i, i+findStrLength).equals(findStr)){
                        count++;
                    }
                }
            }
        }
        System.out.println(count);

Answer 3

Here it is, wrapped up in a nice and reusable method:

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 4

A lot of the given answers fail on one or more of:

• Patterns of arbitrary length
• Overlapping matches (such as counting "232" in "23232" or "aa" in "aaa")
• Regular expression meta-characters

Here's what I wrote:

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;
}

Example call:

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

If you want a non-regular-expression search, just compile your pattern appropriately with the LITERAL flag:

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

Answer 5

Based on the existing answer(s) I'd like to add a "shorter" version without the if:

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

int count = 0, lastIndex = 0;
while((lastIndex = str.indexOf(findStr, lastIndex)) != -1) {
    lastIndex += findStr.length() - 1;
    count++;
}

System.out.println(count); // output: 3

Answer 6

If you need the index of each substring within the original string, you can do something with indexOf like this:

 private static List<Integer> getAllIndexesOfSubstringInString(String fullString, String substring) {
    int pointIndex = 0;
    List<Integer> allOccurences = new ArrayList<Integer>();
    while(fullPdfText.indexOf(substring,pointIndex) >= 0){
       allOccurences.add(fullPdfText.indexOf(substring, pointIndex));
       pointIndex = fullPdfText.indexOf(substring, pointIndex) + substring.length();
    }
    return allOccurences;
}