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 }

 

posted @ 2021-03-12 14:30  北叶青藤  阅读(89)  评论(0)    收藏  举报