博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
leetcode393. UTF-8 Validation
阅读量:6804 次
发布时间:2019-06-26

本文共 4256 字,大约阅读时间需要 14 分钟。

题目要求

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 10xxxxxxGiven 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.

检验整数数组能否构成合法的UTF8编码的序列。UTF8的字节编码规则如下:

  1. 每个UTF8字符包含1~4个字节
  2. 如果只包含1个字节,则该字节以0作为开头,剩下的位随意
  3. 如果包含两个或两个以上字节,则起始字节以n个1和1个0开头,例如,如果该UTF8字符包含两个字节,则第一个字节以110开头,同理,三个字符的第一个字节以1110开头。剩余的字节必须以10开头。

思路和代码

首先我们整理一下,每一种类型的UTF8字符包含什么样的规格:

  1. 只包含一个字节,该字节格式为0xxxxxxx,则转换为整数的话,该整数必须小于128(1000000)
  2. 包含多个字节,则头字节格式为110xxxxx, 1110xxxx, 11110xxx。而紧跟其后的字符必须格式为10xxxxxx。

综上所述:

  1. num<1000000: 单字节
  2. 10000000=<num<11000000: 多字节字符的跟随字节
  3. 11000000<=num<11100000: 两个字节的起始字节
  4. 11100000<=num<11110000: 三个字节的起始字节
  5. 11110000<=num<11111000: 四个字节的起始字节

下面分别是这题的两种实现:

递归实现:

private static final int ONE_BYTE = 128; //10000000    private static final int FOLLOW_BYTE = 192; //11000000    private static final int TWO_BYTE = 224; //11100000    private static final int THREE_BYTE = 240;//11110000    private static final int FOUR_BYTE = 248;//11111000    public boolean validUtf8(int[] data) {        return validUtf8(data, 0);    }        public boolean validUtf8(int[] data, int startAt) {        if(startAt >= data.length) return true;        int first = data[startAt];                int followLength = 0;        if(first < ONE_BYTE) {            return validUtf8(data, startAt+1);        }else if(first < FOLLOW_BYTE){            return false;        }else if(first 
data.length) return false; for(int i = 1 ; i
= FOLLOW_BYTE) { return false; } } return validUtf8(data, startAt + followLength); }

循环实现:

private static final int ONE_BYTE = 128; //10000000    private static final int FOLLOW_BYTE = 192; //11000000    private static final int TWO_BYTE = 224; //11100000    private static final int THREE_BYTE = 240;//11110000    private static final int FOUR_BYTE = 248;//11111000    public boolean validUtf8(int[] data) {        return validUtf8(data, 0);    }        public boolean validUtf8(int[] data, int startAt) {        int followCount = 0;        for(int num : data) {            if(num < ONE_BYTE) {                if(followCount != 0) {                    return false;                }            }else if(num < FOLLOW_BYTE) {                if(followCount == 0) {                    return false;                }                followCount--;            }else if(num < TWO_BYTE) {                if(followCount != 0) {                    return false;                }                followCount = 1;            }else if(num < THREE_BYTE) {                if(followCount != 0) {                    return false;                }                followCount = 2;            }else if(num < FOUR_BYTE) {                if(followCount != 0) {                    return false;                }                followCount = 3;            }else {                return false;            }        }        return followCount == 0;    }

转载地址:http://lknwl.baihongyu.com/

你可能感兴趣的文章
Oracle 10.2.0.5 FOR ALL OS patch官方数据库10GU5补丁 百度云
查看>>
Cocos2d 各种动作介绍
查看>>
Linux 环境下安装 Nginx+php+mysql 开发环境
查看>>
在.NET “.NET研究”4中用IIS部署WCF就这么简单
查看>>
卡巴斯基安全公告:2008恶意软件发展情况
查看>>
IE9 Beta与四大浏览器基准测试对比
查看>>
PAP认证方式原理和实现
查看>>
PDO 和 MySQLi 区别
查看>>
wamp安装及配置
查看>>
引用计数
查看>>
svn -- svn简介
查看>>
常存敬畏之心
查看>>
AgileEAS.NET平台开发Step By Step系列-药店系统-索引
查看>>
基于邻接表的新顶点的增加
查看>>
1118 Birds in Forest (25 分)
查看>>
追逐心目中的那个Ta
查看>>
【Spiral Matrix】cpp
查看>>
使用sphinx索引mysql数据
查看>>
视频连续播放
查看>>
C# 中的委托和事件(转)
查看>>