UTF-8 Validation问题及解法

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问题描述：

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 | 0xxxxxxx0000 0080-0000 07FF | 110xxxxx 10xxxxxx0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx0001 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.

示例:

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.

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.

过程详见代码：

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

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