[LeetCode] 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.
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.

Thought process:
Use right shift.

Solution:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
class Solution {
    public boolean validUtf8(int[] data) {
        int count = 0;
        for (int i : data) {
            if (count == 0) {
                if (i >> 3 == 0b11110) {
                    count = 3;
                } else if (i >> 4 == 0b1110) {
                    count = 2;
                } else if (i >> 5 == 0b110) {
                    count = 1;
                } else if (i >> 7 == 1) {
                    return false;
                }
            } else {
                if (i >> 6 != 0b10) {
                    return false;
                }
                count--;
            }
        }
        return count == 0;
    }
}

Time complexity: O(n).

Comments

Popular posts from this blog

[LeetCode] 714. Best Time to Buy and Sell Stock with Transaction Fee

[LeetCode] 269. Alien Dictionary

[LeetCode] 631. Design Excel Sum Formula