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):

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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();     } }