> For the complete documentation index, see [llms.txt](https://nataliekung.gitbook.io/ladder_code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nataliekung.gitbook.io/ladder_code/meta-2025/408.-valid-word-abbreviation.md).

# 408. Valid Word Abbreviation

A string can be **abbreviated** by replacing any number of **non-adjacent**, **non-empty** substrings with their lengths. The lengths **should not** have leading zeros.

For example, a string such as `"substitution"` could be abbreviated as (but not limited to):

* `"s10n"` (`"s ubstitutio n"`)
* `"sub4u4"` (`"sub stit u tion"`)
* `"12"` (`"substitution"`)
* `"su3i1u2on"` (`"su bst i t u ti on"`)
* `"substitution"` (no substrings replaced)

The following are **not valid** abbreviations:

* `"s55n"` (`"s ubsti tutio n"`, the replaced substrings are adjacent)
* `"s010n"` (has leading zeros)
* `"s0ubstitution"` (replaces an empty substring)

Given a string `word` and an abbreviation `abbr`, return *whether the string **matches** the given abbreviation*.

A **substring** is a contiguous **non-empty** sequence of characters within a string.

 

**Example 1:**

<pre><code><strong>Input: word = "internationalization", abbr = "i12iz4n"
</strong><strong>Output: true
</strong><strong>Explanation: The word "internationalization" can be abbreviated as "i12iz4n" ("i nternational iz atio n").
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: word = "apple", abbr = "a2e"
</strong><strong>Output: false
</strong><strong>Explanation: The word "apple" cannot be abbreviated as "a2e".
</strong></code></pre>

&#x20;

**Constraints:**

* `1 <= word.length <= 20`
* `word` consists of only lowercase English letters.
* `1 <= abbr.length <= 10`
* `abbr` consists of lowercase English letters and digits.
* All the integers in `abbr` will fit in a 32-bit integer.

分析

遇到数字叠加，注意开头0的情况 用NUM==0 AND C == '0'判断

遇到字母， 更新POS比较

注意不要忘了最后的NUMBER!!!!

```
class Solution:
    def validWordAbbreviation(self, word: str, abbr: str) -> bool:
        # Input: word = "internationalization", abbr = "i12iz4n"
        pos = 0
        num = 0
        for i,c in enumerate(abbr):
            if c.isdigit():
                if num == 0 and c == '0':
                    return False
                num = num * 10 + int(c)
            else:
                pos += num
                num = 0
                if pos >= len(word) or word[pos] != c:
                    return False
                pos += 1
        return pos + num == len(word) #别忘了最后的NUMBER！！！！！！！




```