408. Valid Word Abbreviation

• Difficulty: Easy

• Topics: String

• Similar Questions:

  • Minimum Unique Word Abbreviation

  • Word Abbreviation

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. </pre> </p>

Solutions:

class Solution {
public:
    bool validWordAbbreviation(string word, string abbr) {
        int pos = 0;
        int count = 0;
        for (int i = 0; i < abbr.length(); ++i) {
            if (!isdigit(abbr[i])) {
                pos += count;
                count = 0;
                if (pos > word.length() || word[pos] != abbr[i])    return false;
                ++pos;
            } else {
                if (count == 0 && abbr[i] == '0')   return false;
                count = 10 * count + (abbr[i] - '0');
            }
        }

        pos += count;
        return pos == word.length();
    }
};