题目描述:
LeetCode 393. UTF-8 Validation
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
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.
题目大意:
一个UTF8字符的长度从1到4个字节不等,服从下列规则:
对于1字节的字符,首位数是0,后面是unicode代码。
对于n字节的字符,前n位数全是1,第n+1位是0,后面跟着n-1个字节,最高位的两位数是10。
详见上表。
给定一个整数数组表示的数据,判断其是否为有效的utf-8编码。
注意:
输入是整数数组。只有最低位的8位数用来存放数据。这意味着每一个整数只表示一个字节的数据。
解题思路:
模拟题+位运算(Bit Manipulation)
Python代码:
class Solution(object):
def validUtf8(self, data):
"""
:type data: List[int]
:rtype: bool
"""
masks = [0x0, 0x80, 0xE0, 0xF0, 0xF8]
bits = [0x0, 0x0, 0xC0, 0xE0, 0xF0]
while data:
for x in (4, 3, 2, 1, 0):
if data[0] & masks[x] == bits[x]:
break
if x == 0 or len(data) < x:
return False
for y in range(1, x):
if data[y] & 0xC0 != 0x80:
return False
data = data[x:]
return True
本文链接:http://bookshadow.com/weblog/2016/09/04/leetcode-utf-8-validation/
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