Given a non-empty string word 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
Notice that only the above abbreviations are valid abbreviations of the string word. Any other string is not a valid abbreviation of word.
Example
Example 1:
Given s = "internationalization", abbr = "i12iz4n":
Return true.
Example 2:
Given s = "apple", abbr = "a2e": Return false.
Solution.
This problem is fairly simple and straightforward. The only tricky part is all the corner cases.
1. if there is a number in the abbr string, parse it to num.
2. advance the index of word by num.
3. if both i and j are still in bound, check if match or not.
4. repeat the above steps until either i or j is out of bound.
5. check if both i and j are out of bound.
1 public class Solution { 2 public boolean validWordAbbreviation(String word, String abbr) { 3 if(abbr == null || abbr.length() == 0){ 4 return false; 5 } 6 int i = 0; 7 int j = 0; 8 int num = 0; 9 while(i < word.length() && j < abbr.length()){ 10 while(j < abbr.length() && abbr.charAt(j) >= '0' 11 && abbr.charAt(j) <= '9'){ 12 num = num * 10 + (abbr.charAt(j) - '0'); 13 if(num == 0){ 14 return false; 15 } 16 j++; 17 } 18 i += num; 19 num = 0; 20 if(i >= word.length() || j >= abbr.length()){ 21 break; 22 } 23 else if(word.charAt(i) != abbr.charAt(j)){ 24 return false; 25 } 26 else{ 27 i++; 28 j++; 29 } 30 } 31 return i == word.length() && j == abbr.length(); 32 } 33 }
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