UTF-8 Validation
Desicription
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:
代码语言：javascript复制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 10xxxxxxGiven 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:
代码语言：javascript复制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:
代码语言：javascript复制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
代码语言：javascript复制class Solution {
public:
bool validUtf8(const std::vector<int>& data) {
for(int i = 0; i < data.size(); i ) {
if(data[i] >> 7 == 0) {
continue;
} else if(data[i] >> 5 == 0b110) {
for(int count = 1; count <= 1; count ) {
if(count i >= data.size() || data[i count] >> 6 != 0b10) {
return false;
}
}
i = 1;
} else if(data[i] >> 4 == 0b1110) {
for(int count = 1; count <= 2; count ) {
if(count i >= data.size() || data[i count] >> 6 != 0b10) {
return false;
}
}
i = 2;
} else if(data[i] >> 3 == 0b11110) {
for(int count = 1; count <= 3; count ) {
if(count i >= data.size() || data[i count] >> 6 != 0b10) {
return false;
}
}
i = 3;
} else {
return false;
}
}
return true;
}
};
