## 题目描述：

LeetCode 393. UTF-8 Validation

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

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

1. For 1-byte character, the first bit is a 0, followed by its unicode code.
2. 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.
```

## Python代码：

``````class Solution(object):
def validUtf8(self, data):
"""
:type data: List[int]
:rtype: bool
"""
masks = [0x0, 0x80, 0xE0, 0xF0, 0xF8]
bits = [0x0, 0x0, 0xC0, 0xE0, 0xF0]
while data:
for x in (4, 3, 2, 1, 0):
if data & masks[x] == bits[x]:
break
if x == 0 or len(data) < x:
return False
for y in range(1, x):
if data[y] & 0xC0 != 0x80:
return False
data = data[x:]
return True
``````

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