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.

public class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        if(word.length() == 0 && abbr.length()== 0) return true;
        if(word.length() == 0 || abbr.length()== 0) return false;
        if(abbr.charAt(0) == '0') return false; // edge case a -> 01
        int cur = 0;
        int point = 0;
        for(int i = 0 ; i < abbr.length() ; i++){
            char temp = abbr.charAt(i);
            if(Character.isDigit(temp)) {
                cur = cur * 10 + temp - '0';
                if(cur == 0) return false;   //edge case  hi -> h01
            }
            if(temp  >= 'a' && temp <= 'z'){
                point = point + cur;
                if(point >= word.length()) return false;
                if(abbr.charAt(i) == word.charAt(point)){
                    point ++;
                    cur = 0;
                }
                else
                    return false;
            }
        }
       
        if(cur != 0) point = point+ cur;  //edge case "inter" -> "inte1"
        return point == word.length();
    }
}