Leetcode: 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.

这道题题干给出了判断 one single UTF-8 char的方法，然后给一个UTF-8 char sequence，判断是不是正确sequence. (读题读了很久)

这道题关键是要学到用 & 取出一个bit sequence当中几位的方法

二进制数表示法：在前面加 0b, 八进制加0o, 十六进制加0x

 1 public class Solution {
 2     public boolean validUtf8(int[] data) {
 3         if (data==null || data.length==0) return false;
 4         for (int i=0; i<data.length; i++) {
 5             if (data[i] > 255) return false;
 6             int moreChecks = 0; //moreCheck is the number of more bytes that need to check for this char
 7             if ((data[i] & 0b10000000) == 0) moreChecks = 0;
 8             else if ((data[i] & 0b11100000) == 0b11000000) moreChecks = 1;
 9             else if ((data[i] & 0b11110000) == 0b11100000) moreChecks = 2;
10             else if ((data[i] & 0b11111000) == 0b11110000) moreChecks = 3;
11             else return false;
12             for (int j=1; j<=moreChecks; j++) {
13                 if (i+j >= data.length) return false;
14                 if ((data[i+j] & 0b11000000) != 0b10000000) return false;
15             }
16             i = i + moreChecks;
17         }
18         return true;
19     }
20 }