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.

注意：考虑word＝’a‘， abbr＝’01‘， test case return false。因此，所有leading zeros的temp都需要返回false。

class Solution {
public:
    bool validWordAbbreviation(string word, string abbr) {
        int wordIndex = 0, abbrIndex = 0;
        while (wordIndex < word.size()) {
            int abbrIndexEnd = abbrIndex;
            while (abbrIndexEnd < abbr.size() && isdigit(abbr[abbrIndexEnd]))
                abbrIndexEnd++;
            
            // is a char and we need to check whether word[wordIndex] = abbr[abbrIndex]
            if (abbrIndexEnd == abbrIndex) {
                if (word[wordIndex] != abbr[abbrIndex]) return false;
                wordIndex++;
                abbrIndex++;
            } else {
                string temp = abbr.substr(abbrIndex, abbrIndexEnd - abbrIndex);
                if (temp[0] == '0') return false;
                int abbrLength = atoi(temp.c_str());
                wordIndex += abbrLength;
                abbrIndex = abbrIndexEnd;
            }
        }
        return wordIndex == word.size() && abbrIndex == abbr.size();
    }
};