LeetCode 0393 - UTF-8 Validation

Reck Zhang

发布于 2021-08-11 11:05:57

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发布于 2021-08-11 11:05:57

文章被收录于专栏：Reck Zhang

UTF-8 Validation

Desicription

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

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

Solution

class Solution {
public:
    bool validUtf8(const std::vector<int>& data) {
        for(int i = 0; i < data.size(); i++) {
            if(data[i] >> 7 == 0) {
                continue;
            } else if(data[i] >> 5 == 0b110) {
                for(int count = 1; count <= 1; count++) {
                    if(count + i >= data.size() || data[i + count] >> 6 != 0b10) {
                        return false;
                    }
                }
                i += 1;
            } else if(data[i] >> 4 == 0b1110) {
                for(int count = 1; count <= 2; count++) {
                    if(count + i >= data.size() || data[i + count] >> 6 != 0b10) {
                        return false;
                    }
                }
                i += 2;
            } else if(data[i] >> 3 == 0b11110) {
                for(int count = 1; count <= 3; count++) {
                    if(count + i >= data.size() || data[i + count] >> 6 != 0b10) {
                        return false;
                    }
                }
                i += 3;
            } else {
                return false;
            }
        }
        return true;
    }
};

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原始发表：2019-09-06，如有侵权请联系 cloudcommunity@tencent.com 删除

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