[leetcode] 393. UTF-8 Validation 解题报告_393. utf-8 编码验证-优快云博客

本文链接：https://blog.youkuaiyun.com/qq508618087/article/details/52483893

该博客详细介绍了LeetCode上的393题——UTF-8验证的解题过程。作者指出，UTF-8字符根据长度规则可以为1到4个字节，并列举了其编码特点。文章通过举例说明了UTF-8编码的工作原理，并提出通过计数器判断当前字符在编码中的位置来判断整个字符串是否符合UTF-8编码规范。

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题目链接：https://leetcode.com/problems/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.

思路：题目并不难，但却很难写的好看．借助一个计数器来标志当前是否是开始，或者处在一个多于一个bite的unicode的第几个．

代码如下：

class Solution {
public:
    bool validUtf8(vector<int>& data) {
        int count = 0;
        for(const auto &val: data)
        {
            if(!count)
            {
                if((val>>5)==0b110) count = 1;
                else if((val>>4)==0b1110) count = 2;
                else if((val>>3)==0b11110) count = 3;
                else if(val>>7) return false;
                continue;
            }
            if((val>>6)!=0b10) return false;
            count--;
        }
        return true;
    }
};

参考：https://discuss.leetcode.com/topic/57195/concise-c-implementation