393. UTF-8 Validation
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
(hexadecimal) | (binary)
--------------------+---------------------------------------------
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.
Solution
(1) Java
(2) Python
class Solution:
def validUtf8(self, data):
"""
:type data: List[int]
:rtype: bool
"""
nbytes = 0
for i in range(len(data)):
num = data[i]&255
if num&128 == 0:
if nbytes != 0:
return False
else:
continue
elif nbytes == 0 and num&192 == 128:
return False
elif nbytes > 0 and num&192 == 128:
nbytes -= 1
elif nbytes == 0 and num&248==240:
nbytes = 3
elif nbytes == 0 and num&240==224:
nbytes = 2
elif nbytes == 0 and num&224==192:
nbytes = 1
else:
return False
return True if nbytes == 0 else False
(3) Scala