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.
看上去不难,但是这道题坑非常多,而且test case 设计的不太好。
public class Solution { public boolean validWordAbbreviation(String word, String abbr) { int pos = 0; int i = 0; int num = 0; while(i<abbr.length()){ if(Character.isDigit(abbr.charAt(i))){ while(i<abbr.length() && Character.isDigit(abbr.charAt(i))){ if(abbr.charAt(i) == '0'&&num == 0){ return false; } if(num == 0){ num += Integer.parseInt(abbr.charAt(i)+""); } else{ num = num*10 + Integer.parseInt(abbr.charAt(i)+""); } i++; } pos += num; num = 0; } else{ if(pos >= word.length()){ return false; } if(abbr.charAt(i) != word.charAt(pos)){ return false; } pos++; i++; } } return pos == word.length(); } }