# 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.
```

## Analysis

Easy题，但是有许多实现细节容易出问题。

* 用`Character.isDigit(t[j])`来检测是否为数字
* 数字字符到数字的转换
* while嵌套while循环要注意内层循环中的指针不要越界

## Solution

```java
class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        if (word == null || abbr == null) {
            return false;
        }
        int i = 0, j = 0;
        int wordLength = word.length();
        int abbrLength = abbr.length();
        char[] s = word.toCharArray();
        char[] t = abbr.toCharArray();

        while (i < wordLength && j < abbrLength) {
            if (Character.isDigit(t[j])) {
                if (t[j] == '0') {
                    return false;
                }
                int val = 0;
                while (j < abbrLength && Character.isDigit(t[j])) {
                    val = val * 10 + (t[j] - '0');
                    j++;
                }
                i += val;
            } else {
                if (s[i] != t[j]) {
                    return false;
                }
                i++;
                j++;
            }
        }
        return i == wordLength && j == abbrLength;

    }
}
```

## Reference

<https://leetcode.com/problems/valid-word-abbreviation/discuss/89523/Short-and-easy-to-understand-Java-Solution/94182>


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