【leetcode】393. UTF-8 Validation【M】_leedcode 393-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。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Subscribe to see which companies asked this question

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