Leetcode 408. Valid Word Abbreviation

Given a non-empty string s 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"]

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

Note:
Assume s contains only lowercase letters and abbr contains only lowercase letters and digits.

Example 1:

Given s = "internationalization", abbr = "i12iz4n":

Return true.

Example 2:

Given s = "apple", abbr = "a2e":

Return false.

Code (Java):

class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        if (word.isEmpty()) {
            return abbr.isEmpty();
        }

        int pWord = 0;
        int pAbbr = 0;

        while (pWord < word.length() && pAbbr < abbr.length()) {
            if (!Character.isDigit(abbr.charAt(pAbbr))) {
                if (word.charAt(pWord) == abbr.charAt(pAbbr)) {
                    pWord++;
                    pAbbr++;
                } else {
                    return false;
                }
            } else {           
                // edge case: leading zero
                //
                if (abbr.charAt(pAbbr) == '0') {
                    return false;
                }
                
                int num = 0;
                while (pAbbr < abbr.length() && Character.isDigit(abbr.charAt(pAbbr))) {
                    int digit = Character.getNumericValue(abbr.charAt(pAbbr));
                    num = num * 10 + digit;
                    pAbbr++;
                }

                pWord += num;
            }
        }

        return pWord == word.length() && pAbbr == abbr.length();
    }
}