内容简介: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: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:综上所述:
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 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.
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:
- 每个UTF8字符包含1~4个字节
- 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
- 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:
- 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)
- 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:
- num<1000000: 单字节
- 10000000=<num<11000000: 多字节字符的跟随字节
- 11000000<=num<11100000: 两个字节的起始字节
- 11100000<=num<11110000: 三个字节的起始字节
- 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 <TWO_BYTE) {
followLength = 2;
}else if(first < THREE_BYTE) {
followLength = 3;
}else if(first < FOUR_BYTE) {
followLength = 4;
}else {
return false;
}
if(startAt + followLength > data.length) return false;
for(int i = 1 ; i<followLength ; i++) {
int next = data[startAt + i];
if(next < ONE_BYTE || next >= 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;
}
以上所述就是小编给大家介绍的《leetcode393. UTF-8 Validation》,希望对大家有所帮助,如果大家有任何疑问请给我留言,小编会及时回复大家的。在此也非常感谢大家对 码农网 的支持!
猜你喜欢:本站部分资源来源于网络,本站转载出于传递更多信息之目的,版权归原作者或者来源机构所有,如转载稿涉及版权问题,请联系我们。
Effective Objective-C 2.0
Matt Galloway / 爱飞翔 / 机械工业出版社 / 2014-1 / 69.00元
《effective objective-c 2.0:编写高质量ios与os x代码的52个有效方法》是世界级c++开发大师scott meyers亲自担当顾问编辑的“effective software development series”系列丛书中的新作,amazon全五星评价。从语法、接口与api设计、内存管理、框架等7大方面总结和探讨了objective-c编程中52个鲜为人知和容易被忽......一起来看看 《Effective Objective-C 2.0》 这本书的介绍吧!