[lintcode] problem 637 – valid word abbreviation

Given a non-empty string word and an abbreviation abbr, return whether the string matches with the given abbreviation.

A string such as “word” contains only the following valid abbreviations:

[“word”, “1ord”, “w1rd”, “wo1d”, “wor1”, “2rd”, “w2d”, “wo2”, “1o1d”, “1or1”, “w1r1”, “1o2”, “2r1”, “3d”, “w3”, “4”]

Note

Notice that only the above abbreviations are valid abbreviations of the string word. Any other string is not a valid abbreviation of word.

Example

No.1

Input : s = “internationalization”, abbr = “i12iz4n”

Output : true

No.2

Input : s = “apple”, abbr = “a2e”

Output : false

Code

public boolean (String word, String abbr) {
    int idx = 0;
    int count = 0;

    for (int i = 0; i < abbr.length(); i++) {
        char ch = abbr.charAt(i);

        if (Character.isDigit(ch)) {
            if (ch == '0' && count == 0)
                return false;
            
            count = count * 10 + ch - '0';
        }
        else {
            idx += count;

            if (idx >= word.length() || word.charAt(idx) != ch)
                return false;

            idx++;
            count = 0;
        }
    }

    return idx + count == word.length();
}

[lintcode] problem 637 – valid word abbreviation

Note

Example

No.1

No.2

Code

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