393. UTF-8 Validation

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

1. For 1-byte character, the first bit is a 0, followed by its unicode code.
2. 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.

题点就是得到符合格式的least significant 8 bits。
用mask 100000000， 111000000（n=2），111100000（n=3），
     11111000(n=4),110000000(for 10xxxxxx)

   public boolean validUtf8(int[] data) {  
     int mask1 = (int)Math.pow(2,7),   
       mask2 = mask1+(int) Math.pow(2,6)+(int)Math.pow(2,5),  
       mask3 =mask2+(int)Math.pow(2,4),   
       mask4 =mask3+(int)Math.pow(2,3),  
       mask =mask1 +(int) Math.pow(2,6);  
     int n = 0;  
     for(int i = 0; i < data.length; i++){  
       if(n == 0){  
         if((data[i] & mask1) == 0)  
           continue;  
         else if((data[i] & mask2) == mask2 - Math.pow(2,5))  
           n = 1;  
         else if((data[i] & mask3) == mask3 - Math.pow(2,4))  
           n = 2;  
         else if((data[i] & mask4) == mask4 - Math.pow(2,3))  
           n = 3;  
         else return false;  
       }else{  
         if((data[i] & mask) != mask1) return false;  
         n--;  
       }  
     }  
     return n==0;  
   }