We are given a string, say, "itiswhatitis" and a substring, say, "is".
I need to find the index of 'i' when the string "is" occurs a second time in the original string.
String.indexOf("is") will return 2 in this case. I want the output to be 10 in this case.
Use overloaded version of indexOf(), which takes the starting index (fromIndex) as 2nd parameter:
str.indexOf("is", str.indexOf("is") + 1);
I am using: Apache Commons Lang: StringUtils.ordinalIndexOf()
StringUtils.ordinalIndexOf("Java Language", "a", 2)
int first = string.indexOf("is");
int second = string.indexOf("is", first + 1);
This overload starts looking for the substring from the given index.
string.indexOf("is", second + 1); although it would probably be better to make a method that returns the Nth indexOfYou can write a function to return array of occurrence positions, Java has String.regionMatches function which is quite handy
public static ArrayList<Integer> occurrencesPos(String str, String substr) {
final boolean ignoreCase = true;
int substrLength = substr.length();
int strLength = str.length();
ArrayList<Integer> occurrenceArr = new ArrayList<Integer>();
for(int i = 0; i < strLength - substrLength + 1; i++) {
if(str.regionMatches(ignoreCase, i, substr, 0, substrLength)) {
occurrenceArr.add(i);
}
}
return occurrenceArr;
}
if you want to find index for more than 2 occurrence:
public static int ordinalIndexOf(String fullText,String subText,int pos){
if(fullText.contains(subText)){
if(pos <= 1){
return fullText.indexOf(subText);
}else{
--pos;
return fullText.indexOf(subText, ( ordinalIndexOf(fullText,subText,pos) + 1) );
}
}else{
return -1;
}
}
I hope I'm not late to the party.. Here is my answer. I like using Pattern/Matcher because it uses regex which should be more efficient. Yet, I think this answer could be enhanced:
Matcher matcher = Pattern.compile("is").matcher("I think there is a smarter solution, isn't there?");
int numOfOcurrences = 2;
for(int i = 0; i < numOfOcurrences; i++) matcher.find();
System.out.println("Index: " + matcher.start());
It seems to be a good party... I'm in:
public static int nthIndexOf(String str, String subStr, int count) {
int ind = -1;
while(count > 0) {
ind = str.indexOf(subStr, ind + 1);
if(ind == -1) return -1;
count--;
}
return ind;
}
Anyone who is looking for Nth occurance of string
public class NthOccuranceExample {
public static void main(String[] args) {
String str1 = "helloworld good morning good evening good night";
String str2 = "ing";
int n = 2;
int index = nthOccurrence(str1, str2, n);
System.out.println("index of str2 in str1 at occurrence "+ n +" = "+ index);
}
public static int nthOccurrence(String str1, String str2, int n) {
String tempStr = str1;
int tempIndex = -1;
int finalIndex = 0;
for(int occurrence = 0; occurrence < n ; ++occurrence){
tempIndex = tempStr.indexOf(str2);
if(tempIndex==-1){
finalIndex = 0;
break;
}
tempStr = tempStr.substring(++tempIndex);
finalIndex+=tempIndex;
}
return --finalIndex;
}
}
You can get any occurrence of a substring in a string with a recursive method like this without any libraries:
import java.util.Arrays;
public class Test {
public static void main(String[] args) {
String word = "unrestfulness";
String[] temp = new String[word.length()];
for (char c : word.toCharArray()) {
int count = (int) Arrays.stream(temp).filter(e -> e != null && e.contains(String.valueOf(c))).count();
int index = getIndex(word, String.valueOf(c), count);
System.out.println(c + " " + count + " " + index);
temp[index] = String.valueOf(c);
}
System.out.println("result -> " + Arrays.toString(temp));
}
public static int getIndex(String word, String letter, int count) {
return count == 0 ? word.indexOf(letter) : word.indexOf(letter, getIndex(word, letter, count - 1) + 1);
}
}
i think a loop can be used.
1 - check if the last index of substring is not the end of the main string.
2 - take a new substring from the last index of the substring to the last index of the main string and check if it contains the search string
3 - repeat the steps in a loop