393. UTF 8 Validation - jiejackyzhang/leetcode-note GitHub Wiki

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.

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.

解题思路： Thanks to the idea from http://bookshadow.com/weblog/2016/09/04/leetcode-utf-8-validation/

I wrote the Java version.

The basic idea is:

1. First determine if a n-bytes character start with '0', '110', '1110' or '11110' and find out n.
2. Check if the following n-1 bytes start with '10'.
3. Do the same thing to next character until the end of the array data.

public class Solution {
    public boolean validUtf8(int[] data) {
        int[] masks = new int[]{0x0, 0x80, 0xE0, 0xF0, 0xF8};
        int[] bits = new int[]{0x0, 0x0, 0xC0, 0xE0, 0xF0};
        int start = 0;
        while(start < data.length) {
            int n;
            // start from a character
            // determine its number of bytes (n-bytes) 
            for(n = 4; n >= 0; n--) {
                if((data[start] & masks[n]) == bits[n]) break;
            }
            // if not beginning with '0', '110', '1110', '11110'
            // or not have enough length
            // return false
            if(n == 0 || start + n > data.length ) return false;
            for(int j = 1; j < n; j++) {
                // check if followed by n-1 bytes beginning with '10'.
                if((data[start+j] & 0xC0) != 0x80) return false;
            }
            // continue with next character
            start += n;
        }
        return true;
    }
}