393 UTF-8 Validation

393. UTF-8 Validation

1. Question

A character in UTF8 can be from1 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.

2. Implementation

(1) Bit Manipulation

思路: 这题基本就按照题目给的utf-8的特性去验证，我们用一个变量count检查接下来的有多少个byte需要判断其是否以10开头的。如果count是0的话，我们将data[i]右移相应的位数，比如将其右移3位，判断右移后的数是否等于11110，如果是的话说明是一个4byte的character, 注意如果count是0，而一个byte是10000000的话，这不是valid的utf-8格式

class Solution {
    public boolean validUtf8(int[] data) {
        int count = 0;

        for (int oneByte : data) {
            if (count == 0) {
                if ((oneByte >> 3) == 0b11110) {
                    count = 3;
                }
                else if ((oneByte >> 4) == 0b1110) {
                    count = 2;
                }
                else if ((oneByte >> 5) == 0b110) {
                    count = 1;
                }
                else if ((oneByte >> 7) == 0b1) {
                    return false;
                }
            }
            else {
                if ((oneByte >> 6) == 0b10) {
                    --count;
                }
                else {
                    return false;
                }
            }
        }
        return count == 0;
    }
}

3. Time & Space Complexity

Bit Manipulation: 时间复杂度O(N), 空间复杂度O(1)