393. UTF-8 Validation UTF-8 编码验证

# 题目描述：

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.
``````

# 解题方法

``````class Solution(object):
def validUtf8(self, data):
"""
:type data: List[int]
:rtype: bool
"""
cnt = 0
for d in data:
if cnt == 0:
if (d >> 5) == 0b110:
cnt = 1
elif (d >> 4) == 0b1110:
cnt = 2
elif (d >> 3) == 0b11110:
cnt = 3
elif (d >> 7):
return False
else:
if (d >> 6) != 0b10:
return False
cnt -= 1
return cnt == 0
``````

http://www.cnblogs.com/grandyang/p/5847597.html

# 日期

2018 年 10 月 8 日 —— 终于开学了。