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:

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["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:

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Given s = "internationalization", abbr = "i12iz4n":
Return true.

Example 2:

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Given s = "apple", abbr = "a2e":
Return false.

这个题目也比较简单，小心处理就可以了。

  1. 从左到右扫描两个字符串
  2. 如果abbr中遇到数字，提取这个数字，并在word中跳过相应数量的字符；如果是字母，就和word中的比较，如果不一样的话，说明不匹配。
  3. 扫描到word, abbr任一个字符串到达终点为止
  4. 最后，如果两个字符串都达到了终点，说明两个字符串匹配。
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public boolean validWordAbbreviation(String word, String abbr) {
    int i = 0, j = 0;
    while (i < word.length() && j < abbr.length()) {
        if (abbr.charAt(j) == '0')
            return false;
        if (Character.isDigit(abbr.charAt(j))) {
            int end = j;
            while (end < abbr.length() && Character.isDigit(abbr.charAt(end))) {
                end++;
            }
            int count = Integer.parseInt(abbr.substring(j, end));
            i += count;
            j = end;
        } else {
            if (word.charAt(i++) != abbr.charAt(j++))
                return false;
        }
    }
    return i == word.length() && j == abbr.length();
}