[leetcode] 393. UTF-8 Validation

[leetcode] 393. UTF-8 Validation

Description

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.

分析

题目的意思是：给定一个数组，判断数组是否是UTF-8。

这个解法非常的简洁，通过平移的方式判断每个字节的头部是否合法，如果不合法就返回false，合法的话，记录剩下字节的头部是否合法，这里cnt保存需要判断剩下字节的字节数。

代码

class Solution {public: bool validUtf8(vector& data) { int cnt=0; for(int d:data){ if(cnt==0){ if((d>>5)==0b110) cnt=1; else if((d>>4)==0b1110) cnt=2; else if((d>>3)==0b11110) cnt=3; else if(d>>7) return false; }else{ if((d>>6)!=0b10) return false; cnt--; } } return cnt==0; }};

参考文献

​​[LeetCode] UTF-8 Validation 编码验证​​

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。