題目：
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.

思路：
easy 要注意兩點：1。連續的digit要組合為一個number 2。number不能有前導0，且number不為0

演算法：

    public boolean validWordAbbreviation(String word, String abbr) {
        int i = 0, j = 0, start = -1;

        while (i < word
 
.length() && j < abbr.length()) {
            if (Character.isDigit(abbr.charAt(j))) {
                if (start == -1) {
                    start = j;
                    if (abbr.charAt(j) - '0' == 0){
                        return false;
                    }
                }
                if
 
 (j == abbr.length() - 1) {

                    int num = Integer.parseInt(abbr.substring(start, j + 1));
                    i += num;
                }
                j++;
            } else {
                if (start != -1) {

                    int num = Integer.parseInt(abbr.substring(start, j));
                    i += num;
                    start = -1;
                } else {
                    if (word.charAt(i) == abbr.charAt(j)) {
                        i++;
                        j++;
                    } else {
                        return false;
                    }
                }
            }
        }
        if (i == word.length() && j == abbr.length())
            return true;
        else
            return false;
    }