408. Valid Word Abbreviation

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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.

解法1：

string类的题目似乎用指针可以规避掉一些cornercase的处理。这题用两个指针维护当前比较的起始位置。
先比较是否相等，如果相等则都往前进一步。
如果不相等，则考虑abbr是否为数字，把数字parse出来后i相应的比较位置就前进对应的数字个数。
这里要注意的是，数字的开端不能为０.
C++

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Java

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