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 public 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 (!isDigit(abbr.charAt(j))) {
6 if (word.charAt(i) != abbr.charAt(j)) return false;
7 i++;
8 j++;
9 }
10 else {
11 int num = 0;
12 while (j<abbr.length() && isDigit(abbr.charAt(j))) {
13 num = num*10 + (int)(abbr.charAt(j)-'0');
14 if (num == 0) return false; //"001" with '0' at front should return false
15 j++;
16 }
17 i = i + num;
18 }
19 }
20 if (i==word.length() && j==abbr.length()) return true;
21 return false;
22 }
23
24 public boolean isDigit(char c) {
25 if (c>='0' && c<='9') return true;
26 return false;
27 }
28 }