题目地址：https://leetcode-cn.com/problems/valid-word-abbreviation/
题目描述
Given a non-empty string s and an abbreviation abbr, return whether the string matches with the given abbreviation.
Astring 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.
题目大意
给一个非空 字符串 s 和一个单词缩写 abbr ，判断这个缩写是否可以是给定单词的缩写。
解题方法
双指针
一个指针指向abbr，一个指针指向word，根据abbr的数字情况移动word的指针，如果abbr指针指向的不是数字而是字符，那么判断是否和word字符相同。最后两个指针应该会同时移动到各自的结尾。
这个题有个小坑，就是有前导0的数字是非法的，需要判断一下。
C++代码如下：
class Solution {
public:
bool validWordAbbreviation(string word, string abbr) {
int iword = 0;
int iabbr = 0;
while (iword < word.size() && iabbr < abbr.size()) {
int count = 0;
if (count == 0 && abbr[iabbr] == '0')
return false;
while (abbr[iabbr] >= '0' && abbr[iabbr] <= '9') {
count = 10 * count + abbr[iabbr] - '0';
iabbr ++;
}
iword += count;
if ((iword >= word.size() && iabbr != abbr.size()) ||
(iword != word.size() && iabbr >= abbr.size()))
return false;
if (iword == word.size() && iabbr == abbr.size())
return true;
if (word[iword] != abbr[iabbr])
return false;
iword ++;
iabbr ++;
}
return iword == word.size() && iabbr == abbr.size();
}
};
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