LeetCode 408. Valid Word Abbreviation

原题链接在这里：https://leetcode.com/problems/valid-word-abbreviation/

题目：

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.

题解：

Use two points i and j, when j 指向数字时, i就跳过相应数字. 最后看i, j是否同时到尾部.

Time Complexity: O(n), n是word, abbr中较长的length.

Space: O(1).

AC Java:

public class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        if(word == null && abbr == null){
            return true;
        }
        
        if(word == null || abbr == null){
            return false;
        }
        
        int i = 0;
        int j = 0;
        while(i<word.length() && j<abbr.length()){
            if(word.charAt(i) == abbr.charAt(j)){
                i++;
                j++;
                continue;
            }
            if(abbr.charAt(j)<='0' || abbr.charAt(j)>'9'){
                return false;
            }
            int s = j;
            while(j<abbr.length() && abbr.charAt(j)>='0' && abbr.charAt(j)<='9'){
                j++;
            }
            int jump = Integer.valueOf(abbr.substring(s, j));
            i += jump;
        }
        return i == word.length() && j == abbr.length();
    }
}

跟上Word Abbreviation.