[Leetcode] 408. Valid Word Abbreviation 解题报告

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.

思路：

一道练手题目，注意边界控制即可。

代码：

class Solution {
public:
bool validWordAbbreviation(string word, string abbr) {
int index1 = 0, index2 = 0;
while(index1 < word.length() && index2 < abbr.length()){
if(isdigit(abbr[index2])) {
int value = abbr[index2++] - '0';
if (value == 0) {
return false;
}
while(index2 < abbr.length() && isdigit(abbr[index2])) {
value = 10 * value + abbr[index2++] - '0';
}
index1 += value;
}
else {
if(word[index1++] != abbr[index2++]) {
return false;
}
}
}
if(index1 != word.length() || index2 != abbr.length()) {
return false;
}
return true;
}
};