leetcode 393. UTF-8 Validation 编码判断问题

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.

这道题就是对UTF-8的编码做一个判断，看起来很复杂，其实很简单，只要遍历一次即可，这个用到了位运算，右移判断即可

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>using namespace std;/* 0xxxxxxx 110xxxxx 10xxxxxx 1110xxxx 10xxxxxx 10xxxxxx 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx*/class Solution{public:    bool validUtf8(vector<int>& data)     {        int count = 0;        for (int one : data)        {            if (count == 0)            {                if ((one >> 3) == 0b11110)                    count = 3;                else if ((one >> 4) == 0b1110)                    count = 2;                else if ((one >> 5) == 0b110)                    count = 1;                else if ((one >> 7) == 0b0)                    count = 0;                else                    return false;            }            else            {                if ((one >> 6) == 0b10)                    count--;                else                    return false;            }        }        return count == 0 ? true : false;    }};