Problem:
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.
思路：
Solution (C++):
bool validUtf8(vector& data) {
int count = 0;
for (auto c : data) {
if (count == 0) {
if ((c >> 5) == 0b110) count = 1;
else if ((c >> 4) == 0b1110) count = 2;
else if ((c >> 3) == 0b11110) count = 3;
else if ((c >> 7) != 0) return false;
} else {
if ((c >> 6) != 0b10) return false;
--count;
}
}
return count == 0;
}
性能：
Runtime: ms Memory Usage: MB
思路：
Solution (C++):
性能：
Runtime: ms Memory Usage: MB