[코테] 리트코드 비트 조작 393. UTF-8 Validation

393. UTF-8 Validation

문제 링크

https://leetcode.com/problems/utf-8-validation/description/

문제 설명

Given an integer array data representing the data, return whether it is a valid UTF-8 encoding (i.e. it translates to a sequence of valid UTF-8 encoded characters).

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

For a 1-byte character, the first bit is a 0, followed by its Unicode code. For an n-bytes character, the first n bits are all one’s, the n + 1 bit is 0, followed by n - 1 bytes with the most significant 2 bits being 10. This is how the UTF-8 encoding would work:

x denotes a bit in the binary form of a byte that may be either 0 or 1.

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.

제한 사항

• 1 <= data.length <= 2 * 10^4
• 0 <= data[i] <= 255

입출력 예 #1

1. Input : data = [197,130,1]
2. Output : true
3. Explanation : data represents the octet sequence: 11000101 10000010 00000001. It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.

입출력 예 #2

1. Input : data = [235,140,4]
2. Output : false
3. Explanation : data represented the octet sequence: 11101011 10001100 00000100. 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.

문제 풀이

class Solution:
    def validUtf8(self, data: List[int]) -> bool:
        # 문자 바이트만큼 10으로 시작 판별
        def check(size):
            for i in range(start + 1, start + size + 1):
                if i >= len(data) or (data[i] >> 6) != 0b10:
                    return False
            return True

        start = 0
        while start < len(data):
            # 첫 바이트 기준 총 문자 바이트 판별
            first = data[start]
            if (first >> 3) == 0b11110 and check(3):
                start += 4
            elif (first >> 4) == 0b1110 and check(2):
                start += 3
            elif (first >> 5) == 0b110 and check(1):
                start += 2
            elif (first >> 7) == 0 :
                start += 1
            else : 
                return False
            
        return True