[LeetCode] 408. Valid Word Abbreviation

zone 发布于2019-08-16 13:44 / 991人阅读

Problem

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.

Solution

class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        if (word == null || abbr == null) return false;
        int i = 0, j = 0;
        while (i < word.length() && j < abbr.length()) {
            char ch1 = word.charAt(i), ch2 = abbr.charAt(j);
            if (ch1 == ch2) {
                i++;
                j++;
            } else if (ch2 >= "1" && ch2 <= "9") {
                int start = j;
                while (j < abbr.length() && Character.isDigit(abbr.charAt(j))) j++;
                int count = Integer.valueOf(abbr.substring(start, j));
                i += count;
            } else return false;
        }
        return i == word.length() && j == abbr.length();
    }
}

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408. Valid Word Abbreviation

Given s = internationalization, abbr = i12iz4n: Return true. Given s = apple, abbr = a2e: Return false. public class Solution { public boolean validWordAbbreviation(String word, String abbr) { ...

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