393. UTF-8 Validation.c

/*
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.
*/

bool validUtf8(int* data, int dataSize) {
    int i, n;
    unsigned char k;
    
    n = 0;
    for (i = 0; i < dataSize; i ++) {
        k = data[i] & 0xff;
        if (n) {
            if ((k & 0xc0) != 0x80) {
                return false;
            }
            n --;
        } else {
            while (k & 0x80) {
                k = k << 1;
                n ++;
            }
            if (n) {
                if (n == 1 || n > 4) return false;
                n --;
            }
        }
    }
    if (n) return false;
    return true;
}


/*
Difficulty:Medium
Total Accepted:15.4K
Total Submissions:44.3K


Companies Google
Related Topics Bit Manipulation

*/