[算法分析与设计] leetcode 每周一题: UTF-8 Validation

题目链接：https://leetcode.com/problems/utf-8-validation/discuss/

题目：

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.

思路：

这道题本身并不难，就直接按图索骥，根据第一个字节前几位，来判断后面有几个字节应该是以01开头的bytes， 唯一忘记了的就是C和C++可以用0bXXXX来表示二进制数，因为答案写的比我好，也比我简单，所以就直接上答案了

代码：

class Solution {
public:
    bool validUtf8(vector<int>& data) {
        int count = 0;
        for(auto b : data) {
            if(count == 0) {
                if((b >> 5) == 0b110) count = 1;
                else if((b >> 4) == 0b1110) count = 2;
                else if((b >> 3) == 0b11110) count = 3;
                else if((b >> 7) == 0b1) return false;
            } else {
                if((b >> 6) != 0b10) return false; 
                count --;
            }
        }
        return count == 0;
    }
};