408. Valid Word Abbreviation

Problem:

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.

Analysis:
这道题的坑爹程度也只是稍微比isNumber Valid好点儿。注意两点
1. 数如果是有多位要考虑进去比如12
2. 如果遇到0，直接返回false.

Solution:

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class Solution { public boolean validWordAbbreviation(String word, String abbr) { if (word == null) return abbr == null; int i = 0, j = 0; while (i < word.length() && j < abbr.length()) { char a = word.charAt(i), b = abbr.charAt(j); if (a == b) { i++; j++; } else if (Character.isDigit(b)) { if (b == '0') return false; int num = 0; while (j < abbr.length() && !Character.isLetter(abbr.charAt(j))) { num = num*10 + (abbr.charAt(j) - '0'); j++; } for (int k = 0; k < num; k++) { i++; } } else { return false; } } return i == word.length() && j == abbr.length(); } }