Java String contains "splitted" String

Question

Which is the best way to check if a Java String A contains another String B, even if B chars are not contiguous?

For example: "takaderoka" contains "tkdr" but not "tkkr".

Is there a built-in function, or do I have to write my own? Thank you.

@clcto `str.matches( ".*t.*k.*d.*r.*" )` since `matches` *auto-anchors* with `^` and `$`. — sp00m, Jul 30 '14 at 14:49
@clcto and sp00m With your solution I have to process someway string B. It could be a good solution anyway. Thank you — Sebastian Ikaros Rizzo, Jul 30 '14 at 14:54
The regex everyone is posting will work, but if you want to have some fun (and waste time) look into dynamic programming for finding the longest common subsequence of 2 strings :) — radar, Jul 30 '14 at 14:54

score 1 · Answer 1 · answered Jul 30 '14 at 14:52

1

No built-in function, but can be done on one line:

"takaderoka".matches("tkkr".replace("", ".*"));

Or put it in a function:

static boolean matcher(String one, String two) {
    return one.matches(two.replace("", ".*"));
}

answered Jul 30 '14 at 14:52

blgt

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@user3218114 Yeah, this regex preserves the order... your test will return `true`, as it should – blgt Jul 30 '14 at 15:36

Braj · Answer 2 · 2014-07-30T15:07:09.620

A simple java program that you can use as utility method. [order is also considered.]

public static boolean isMatched(String s1, String s2) {
    int index = -1;
    for (char ch : s2.toCharArray()) {
        if ((index = s1.indexOf(ch, index + 1)) == -1) {
            return false;
        }
    }
    return true;
}

...
System.out.println(isMatched("takaderoka", "tkdr")); // true
System.out.println(isMatched("takaderoka", "tkkr")); // false

you can make it more optimized:

public static boolean isMatched(String s1, String s2) {
    if ((s1.length() > 0 && s2.length() == 0) || (s2.length() > s1.length())) {
        return false;
    } else if (s1.indexOf(s2) != -1 || s1.equals(s2)) {
        return true;
    } else {
        int index = -1;
        for (char ch : s2.toCharArray()) {
            if ((index = s1.indexOf(ch, index + 1)) == -1) {
                return false;
            }
        }
    }
    return true;
}

score 0 · Answer 3 · answered Jul 30 '14 at 14:47

0

You could do a regular expression matching on your search string with ".*" inserted between each character, alternatively just write your own searching for each character in turn.

answered Jul 30 '14 at 14:47

Tim B

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score 0 · Answer 4 · answered Jul 30 '14 at 14:48

0

There is contains() method for the String class which you can use. Check out the method for the String class and also Regex expressions in Java.

answered Jul 30 '14 at 14:48

MustSeeMelons

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score 0 · Answer 5 · answered Jul 30 '14 at 14:56

I'd use apache commons StringUtils.getLevenshteinDistance() function.

Basically, it measures how many modifications you have to make to transform one string into another. So your problem is quite the same as knowing if transforming all the other letters into nulls is the same as the difference between their lengths.

for example

StringUtils.getLevenshteinDistance("takaderoka","tkdr"); //6,   
StringUtils.getLevenshteinDistance("takaderoka","tkkr"); //7,

if the result = "takaderoka".length - "tkdr".length = 6, then "takaderoka" contains "tkdr"

Taras Velykyy · Answer 6 · 2014-07-30T15:12:19.373

You could this method. Could be refactored to have better code.

 private boolean contains(String source, String toMatch) {
    boolean result = false;
    if (toMatch.length() == 0 || source.length() == 0) {
        return result;
    }
    int j = 0;
    for (int i = 0; i < source.length(); i++) {
        if (toMatch.length() - j > source.length() - i) {
            break;
        }
        if (source.charAt(i) == toMatch.charAt(j)) {
            j++;
        }
        if (j == toMatch.length()) {
            result = true;
            break;
        }
    }

    return result;
}

Java String contains "splitted" String

6 Answers6