【leetcode】393. UTF-8 Validation【M】_[leetcode utf-8 validation 编码验证 python-优快云博客

本文介绍了一种判断整数数组是否符合UTF-8编码规则的方法。UTF-8字符长度为1到4字节，遵循特定的编码规则。文章提供了一个Python实现的例子，包括如何解析每个字节并验证其有效性。

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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:

   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.

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class Solution(object):
    #数 最开始有多少个零
    def countNumOfOne(self,s):
        res = 0
        for i in s:
            if i != '1':
                break
            res += 1
        return res

    #把字符串补全八位
    def resize(self,s):
        s = '0' * (8-len(s)) + s
        return s

    def validUtf8(self, data):
        j = 1

        i = self.resize(bin(data[0])[2:])

        l = self.countNumOfOne(i) - 1
        if len(data) == 1:
            return l == -1
        if l == -1:
            l = 0
        # print l
        # print i,' ',l

        while j < len(data):
            i = self.resize(bin(data[j])[2:])
            temp_len = self.countNumOfOne(i)
            # print i,temp_len,l

            #l>1,说明后面还应该有,所以如果当前字节不是一位的，就错了
            if l >= 1:
                if temp_len != 1:
                    return False
                l -= 1
            #如果l==0，说明之前的多位的已经结束了，如果当前还是一个1开头，那就错了
            elif l == 0:
                if temp_len == 1:
                    return False
                l = temp_len - 1
                if l == -1:
                    l = 0
            else:
                if temp_len != 0:
                    return False
                else:
                    l = 0