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.
1 class Solution { 2 public boolean validWordAbbreviation(String word, String abbr) { 3 int i = 0, j = 0; 4 while (i < word.length() && j < abbr.length()) { 5 if (word.charAt(i) == abbr.charAt(j)) { 6 ++i;++j; 7 continue; 8 }
// this is executed only when the charactoers are different. 9 if (abbr.charAt(j) <= '0' || abbr.charAt(j) > '9') { // "a", "01". If number appears ftft, it cannot begin with 0; 10 return false; 11 } 12 int start = j; 13 while (j < abbr.length() && abbr.charAt(j) >= '0' && abbr.charAt(j) <= '9') { 14 ++j; 15 } 16 int num = Integer.valueOf(abbr.substring(start, j)); 17 i += num; 18 } 19 return i == word.length() && j == abbr.length(); 20 } 21 }

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