408. Valid Word Abbreviation

408. 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.

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分三种情况：
1. 如果是数字，吃数字
2. 如果count == 0, 说明大家都指向字符。那就对比字符
3. 如果不是数字，j += count。 此时i, j都会指向字符或超出范围，两者对比会交给下一次的循环。

注意几个corner case：
1. 数字开头为0: a0b
2. 末尾是数字: ab1。 最后检测长度的时候加上count

class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        int i = 0, j = 0;
        int count = 0;
        
        while (i < abbr.length() && j < word.length()) {
            char c = abbr.charAt(i);
            if (Character.isDigit(c)) {
                if (count == 0 && c <= '0') return false;
                count = Character.getNumericValue(c) + count * 10;
                i++;
            }else if (count == 0) {
                if (c != word.charAt(j)) return false;
                i++;
                j++;
            }else {
                j += count;
                count = 0;
            }
        }
        
        return i == abbr.length() && j + count == word.length();
    }
}