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LeetCode Valid Word Abbreviation

2024-09-14 00:50:17   213人阅读

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

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：

双指针，当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.

AC Java:

 1 public class Solution {
 2     public boolean validWordAbbreviation(String word, String abbr) {
 3         if(word == null && abbr == null){
 4             return true;
 5         }
 6         
 7         if(word == null || abbr == null){
 8             return false;
 9         }
10         
11         int i = 0;
12         int j = 0;
13         while(i<word.length() && j<abbr.length()){
14             if(word.charAt(i) == abbr.charAt(j)){
15                 i++;
16                 j++;
17                 continue;
18             }
19             if(abbr.charAt(j)<=‘0‘ || abbr.charAt(j)>‘9‘){
20                 return false;
21             }
22             int s = j;
23             while(j<abbr.length() && abbr.charAt(j)>=‘0‘ && abbr.charAt(j)<=‘9‘){
24                 j++;
25             }
26             int jump = Integer.valueOf(abbr.substring(s, j));
27             i += jump;
28         }
29         return i == word.length() && j == abbr.length();
30     }
31 }

 

LeetCode Valid Word Abbreviation


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