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.

一刷
题解：
input为一个byte一个byte的十进制表达。
utf-8的特性有:

1. 长度可以从1byte到4bytes
2. 对于1byte的数据，其比特位的最高位是0
3. 对于长度2-4byte的数据，其首个byte的高n位全部为1，第n+1位为0，随后的n-1个byte，其高2位全部为10
   首先利用第一个byte判断数据的长度。根据长度判断后面的byte需要出现几次10XXXXX

class Solution {
    public boolean validUtf8(int[] data) {
        int count = 0;
        for(int num : data){
            if(count == 0){
                if((num>>5) == 0b110) count = 1;//2 byte
                else if((num>>4) == 0b1110) count = 2;//3 byte;
                else if((num>>3) == 0b11110) count = 3;//4 byte
                else if((num>>7) == 1) return false;//count==0, occupy one byte
            }else{
                if((num>>6)!=0b10) return false;
                count--;
            }
        }
        return count == 0;
    }
}