LintCode 637: Valid Word Abbreviation

637. Valid Word Abbreviation
     中文English
     Given a non-empty string word 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”]
Example
Example 1:

Input : s = “internationalization”, abbr = “i12iz4n”
Output : true
Example 2:

Input : s = “apple”, abbr = “a2e”
Output : false
Notice
Notice that only the above abbreviations are valid abbreviations of the string word. Any other string is not a valid abbreviation of word.

解法1：双指针
代码如下：

class Solution {
public:/*** @param word: a non-empty string* @param abbr: an abbreviation* @return: true if string matches with the given abbr or false*/bool validWordAbbreviation(string &word, string &abbr) {int nw = word.size();int na = abbr.size();if (nw == 0 && na == 0) return true;if (nw < na) return false;int pw= 0, pa = 0;while (pw < nw && pa < na) {int num = 0;while (abbr[pa] >= '0' && abbr[pa] <= '9'){num = num * 10 + abbr[pa] - '0';pa++;}if (num == 0) {if (word[pw] == abbr[pa]) {pw++; pa++;} else {return false;}} else {pw += num;}   }if (pw == nw && pa == na) return true;return false;}
};

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