# LeetCode – 408. 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.
```
看上去不难，但是这道题坑非常多，而且test case 设计的不太好。

```public class Solution {
public boolean validWordAbbreviation(String word, String abbr) {
int pos = 0;
int i = 0;
int num = 0;
while(i<abbr.length()){
if(Character.isDigit(abbr.charAt(i))){
while(i<abbr.length() && Character.isDigit(abbr.charAt(i))){
if(abbr.charAt(i) == '0'&&num == 0){
return false;
}
if(num == 0){
num += Integer.parseInt(abbr.charAt(i)+"");
}
else{
num = num*10 + Integer.parseInt(abbr.charAt(i)+"");
}
i++;
}
pos += num;
num = 0;

}
else{
if(pos >= word.length()){
return false;
}
if(abbr.charAt(i) != word.charAt(pos)){
return false;
}
pos++;
i++;
}
}
return pos == word.length();
}
}```

[0人评了分，平均: 0/5]