# LeetCode 393. UTF-8 Validation

## Description

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one’s, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:

```   Char. number range  |        UTF-8 octet sequence
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
```

Given an array of integers representing the data, return whether it is a valid utf-8 encoding.

Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.

Example 1:

data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.

Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:

data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.

Return false.
The first 3 bits are all one’s and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that’s correct.
But the second continuation byte does not start with 10, so it is invalid.

## 描述

UTF-8 中的一个字符可能的长度为 1 到 4 字节，遵循以下的规则：

```   Char. number range  |        UTF-8 octet sequence
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
```

data = [197, 130, 1], 表示 8 位的序列: 11000101 10000010 00000001.

data = [235, 140, 4], 表示 8 位的序列: 11101011 10001100 00000100.

### 思路

• 根据条件判断。
1. 如果是 0 开头，表示一个单独的码；2. 如果是 110 开头，则表示后面跟着一个 10 开头的八位字节码；3. 如果是 1110 开头，表示后面跟着两个 10 开头的八位字节码；4. 如果是 11110 开头，表示后面跟着 3 个 10 开头的字节码；
• 所以我们需要判断的是 1. 数字以二进制表示的第一个 0 前面 1 的个数；2. 以 10 开头的个数；
• 我们将数字用二进制表示，然后转换成为字符串，用正则进行匹配。
```# -*- coding: utf-8 -*-
# @Author:             何睿
# @Create Date:        2019-08-24 10:55:33

import re
from typing import List

class Solution:
def validUtf8(self, data: List[int]) -> bool:

next_ = 0
pattern = re.compile("0b(.*?)0.*") # 匹配第一个 0 前面 1 的个数
while next_ != -1 and next_ < len(data):
next_ = self.__check(next_, data, pattern)

return False if next_ == -1 else True

def __check(self, start, data, pattern):

num = data[start]
if not num & 0b10000000: # 以 0 开头
return start + 1
elif num & 0b11000000 == 0b10000000: # 以 10 开头
return -1
else:
match = pattern.match(str(bin(num))) # 正则匹配
if not match: # 如果匹配失败，返回 -1
return -1
length = match.span(1)[1] - match.span(1)[0] # 后面应该跟 10 开头的数的个数
end = start + length  # 当前位置向后移动 length 个位置
if length > 4 or end > len(data): # 如果个数大于4，如果向后挪动 length 个位置后越界
return -1
if all(map(lambda i: data[i] & 0b11000000 == 0b10000000, range(start + 1, end))): # 范围内的数子都需要以 10 开头
return start + length
return -1
```