CODEWITHSUNDEEP
c#javaregex
NAMO
NAMO

Reputation: 273

finding middle character in string using regex only

How can I find middle character with regex only

For example,this shows the expected output

Hello -> l

world -> r

merged -> rg (see this for even number of occurances)

hi -> hi

I -> I

I tried 

(?<=\w+).(?=\w+)

Upvotes: 5

Views: 1902

Answers (4)

bobble bubble
bobble bubble

Reputation: 18490

Related: How to match the middle character in a string with regex?

The following regex is based on @jaytea's approach and works well with e.g. PCRE, Java or C#.

^(?:.(?=.+?(.\1?$)))*?(^..?$|..?(?=\1$))

Here is the demo at regex101 and a .NET demo at RegexPlanet (click the green ".NET" button)

Middle character(s) will be found in the second capturing group. The goal is to capture two middle characters if there is an even amount of characters, else one. It works by a growing capture towards the end (first group) while lazily going through the string until it ends with the captured substring that grows with each repitition. ^..?$ is used to match strings with one or two characters length.

This "growing" works with capturing inside a repeated lookahead by placing an optional reference to the same group together with a freshly captured character into that group (further reading here).

A PCRE-variant with \K to reset and full matches: ^(?:.(?=.+?(.\1?$)))+?\K..?(?=\1$)|^..?

Curious about the "easy solution using balancing groups" that @Qtax mentions in his question.

Upvotes: 0

shreyansh jogi
shreyansh jogi

Reputation: 2102

public static void main(String[] args) {
        String s = "jogijogi";
        int size = s.length() / 2;
        String temp = "";
        if (s.length() % 2 == 0) {
            temp = s.substring(size - 1, (s.length() - size) + 1);
        } else if (s.length() % 2 != 0) {
            temp = s.substring(size, (s.length() - size));
        } else {
            temp = s.substring(1);

        }

        System.out.println(temp);
    }

Upvotes: 1

Veer Shrivastav
Veer Shrivastav

Reputation: 5496

String str="Hello";
String mid="";
int len = str.length();
if(len%2==1)
    mid= Character.toString(str.getCharAt(len/2));
else
    mid= Character.toString(str.getChatAt(len/2))+ Character.toStringstr.getCharAt((len/2)-1));

This should probably work.

Upvotes: 1

Pieter van Ginkel
Pieter van Ginkel

Reputation: 29632

Regular expressions cannot count in the way that you are looking for. This looks like something regular expressions cannot accomplish. I suggest writing code to solve this.

Upvotes: 5

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