leetcode UTF-8 Validation

最新推荐文章于 2022-04-15 21:33:43 发布

ithiker

最新推荐文章于 2022-04-15 21:33:43 发布

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分类专栏： leetcode 文章标签： UTF-8 Validation

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本文介绍了一种有效的UTF-8编码验证算法。通过解析比特位来判断输入数据是否符合UTF-8编码规则，适用于由整数数组表示的UTF-8编码数据验证。

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UTF-8 Validation

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.

My O(n) solution:

class Solution {
public:

    int getFirstOne(const bitset<8>& a) {
        int count = 0;
        int i = 7;
        while (i >=0 && a[i]) {
            count++;
            i--;
        }
        
        return count;
    }
    bool validUtf8(vector<int>& data) {
        for (int i = 0; i < data.size(); ) {
            bitset<8> a(data[i] & 0x00ff);
            int j = getFirstOne(a);
            if (j == 1 || j >= 5) return false;
            for (int k = 1; k < j; ++k) {
                bitset<8> b(data[i+k] & 0x00ff);
                if (getFirstOne(b) != 1) return false;
            }
            j == 0 ? i += 1 : i += j;
        }
        
        return true;
        
    }
};