題目描述：

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.

題目大意

判斷一個字元的編碼（不是一組）是不是有效的UTF-8編碼。規則是：

對於UTF-8編碼中的任意位元組B，如果B的第一位為0，則B獨立的表示一個字元(ASCII碼)； 如果B的第一位為1，第二位為0，則B為一個多位元組字元中的一個位元組(非ASCII字元)； 如果B的前兩位為1，第三位為0，則B為兩個位元組表示的字元中的第一個位元組； 如果B的前三位為1，第四位為0，則B為三個位元組表示的字元中的第一個位元組； 如果B的前四位為1，第五位為0，則B為四個位元組表示的字元中的第一個位元組；

解題方法

注意題目讓判斷的是一個字元，這樣就簡單了很多。方法是使用位運算，首先判斷首字元中的起始位置是什麼，來知道後面跟著幾個字元或者整個字元是個單獨的字元，然後判斷後面跟著的字元是不是都是以01開頭的，個數是不是和第一個字元指示我們的相等。

時間複雜度是O(N)，空間複雜度是O(1)。N是給出的資料的長度。

class Solution(object):
    def validUtf8(self, data):
        """
        :type data: List[int]
        :rtype: bool
        """
        cnt = 0
        for d in data:
            if cnt == 0:
                if (d >> 5) == 0b110:
                    cnt = 1
                elif (d >> 4) == 0b1110:
                    cnt = 2
                elif (d >> 3) == 0b11110:
                    cnt = 3
                elif (d >> 7):
                    return False
            else:
                if (d >> 6) != 0b10:
                    return False
                cnt -= 1
        return cnt == 0

參考資料：

日期

2018 年 10 月 8 日 —— 終於開學了。