[LintCode] Check Word Abbreviation

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.

public class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        if(abbr == null || abbr.length() == 0){
            return false;
        }
        int i = 0;
        int j = 0;
        int num = 0;
        while(i < word.length() && j < abbr.length()){
            while(j < abbr.length() && abbr.charAt(j) >= '0' 
                    && abbr.charAt(j) <= '9'){
                num = num * 10 + (abbr.charAt(j) - '0');   
                if(num == 0){
                    return false;
                }
                j++;
            }
            i += num;
            num = 0;
            if(i >= word.length() || j >= abbr.length()){
                break;
            }
            else if(word.charAt(i) != abbr.charAt(j)){
                return false;    
            }
            else{
                i++;
                j++;
            }
        }
        return i == word.length() && j == abbr.length();
    }
}

 

 

Related Problems

Word Abbreviation