393. UTF-8 Validation

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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:

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.

题意：就是给定一个数组，判断是否是utf-8编码。

思路：这题有几个要注意的地方，首先是数组是整数的，且只保留后8位，所以我开始是还用位运算来把所有数据处理了一遍，只留最后8位，后来看到别人的做法才发现可以那么简单，还有就是这是允许多个utf-8编码组合的，这个要注意，这题的关键在于如何判断多个前缀，以及组合的判断。

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

这段代码用右移代替了一位位的判断，其实也可以转化为string，用startwith来判断，count用抵消的方式也很方便。

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