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

Solution：TwoPointers對比Check, 跳過Abbr

思路： i, j 分別遍歷word 和 abbr，若元素相同，繼續(xù)；若abbr開始出現(xiàn)數(shù)字num，則i跳過i += num，判斷最后是否分別都到達串尾。
Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

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();
    }
}