Given anon-emptystringsand an abbreviationabbr, 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:
Assumescontains only lowercase letters andabbrcontains only lowercase letters and digits.

Example 1:

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

Return true.

Example 2:

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

Return false.

class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        int i = 0, j = 0;
        while (i < word.length() && j < abbr.length()) {
            if (word.charAt(i) == abbr.charAt(j)) {
                ++i;++j;
                continue;
            }
            if (abbr.charAt(j) <= '0' || abbr.charAt(j) > '9') {
                return false;
            }
            int start = j;
            while (j < abbr.length() && abbr.charAt(j) >= '0' && abbr.charAt(j) <= '9') {
                ++j;
            }
            int num = Integer.valueOf(abbr.substring(start, j));
            i += num;
        }
        return i == word.length() && j == abbr.length();
    }
}