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  • Leetcode: 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.
     1 public class Solution {
     2     public boolean validWordAbbreviation(String word, String abbr) {
     3         int i = 0, j = 0;
     4         while (i<word.length() && j<abbr.length()) {
     5             if (!isDigit(abbr.charAt(j))) {
     6                 if (word.charAt(i) != abbr.charAt(j)) return false;
     7                 i++;
     8                 j++;
     9             }
    10             else {
    11                 int num = 0;
    12                 while (j<abbr.length() && isDigit(abbr.charAt(j))) {
    13                     num = num*10 + (int)(abbr.charAt(j)-'0');
    14                     if (num == 0) return false; //"001" with '0' at front should return false
    15                     j++;
    16                 }
    17                 i = i + num;
    18             }
    19         }
    20         if (i==word.length() && j==abbr.length()) return true;
    21         return false;
    22     }
    23     
    24     public boolean isDigit(char c) {
    25         if (c>='0' && c<='9') return true;
    26         return false;
    27     }
    28 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/6194134.html
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