共计 3285 个字符，预计需要花费 9 分钟才能阅读完成。
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;
}