LeetCode #393 - UTF-8 Validation

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本文介绍了一种用于验证UTF-8编码有效性的算法。通过分析输入整数数组，算法检查每个字节是否符合UTF-8编码规则，包括1到4字节长的字符编码规范。文章详细解释了如何通过位操作来判断数据是否正确编码。

题目描述：

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.

//对于UTF-8编码中的任意字节B，如果B的第一位为0，则B独立的表示一个字符(ASCII码)；

//如果B的第一位为1，第二位为0，则B为一个多字节字符中的一个字节(非ASCII字符)；

//如果B的前两位为1，第三位为0，则B为两个字节表示的字符中的第一个字节；

//如果B的前三位为1，第四位为0，则B为三个字节表示的字符中的第一个字节；

//如果B的前四位为1，第五位为0，则B为四个字节表示的字符中的第一个字节；

class Solution {
public:
    bool validUtf8(vector<int>& data) {
        int i=0;
        while(i<data.size())
        {
            int x=0b10000000;
            if((data[i]&x)==0) 
            {
                i++;
                continue;
            }
            int count=0;
            while((data[i]&x)>0)//位操作优先级低，必须加上括号
            {
                count++;
                x=x>>1;
            }
            if(count==1||count>4)  return false;
            for(int j=i+1;j<i+count;j++) //判断指示字节之后的(count-1)个字节是否以10开头
                if((data[j]&0b10000000)==0||(data[j]&0b01000000)!=0) return false;
            i+=count;//count包含了开头的第一个指示字节
        }
        return true;
    }
};