[LeetCode] 408. Valid Word Abbreviation

最新推荐文章于 2025-07-09 08:00:00 发布

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最新推荐文章于 2025-07-09 08:00:00 发布

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原文链接：https://segmentfault.com/a/1190000017047919

本文探讨了如何判断一个非空字符串是否与给定的缩写相匹配的问题，通过一个具体的解决方案，展示了如何使用双指针技巧遍历字符串和缩写，特别关注于缩写的合法性和数字处理。

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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();
    }
}