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 | 0xxxxxxx0000 0080-0000 07FF | 110xxxxx 10xxxxxx0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx0001 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：

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; //10000000private static final int FOLLOW_BYTE = 192; //11000000private static final int TWO_BYTE = 224; //11100000private static final int THREE_BYTE = 240;//11110000private static final int FOUR_BYTE = 248;//11111000public 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; //10000000private static final int FOLLOW_BYTE = 192; //11000000private static final int TWO_BYTE = 224; //11100000private static final int THREE_BYTE = 240;//11110000private static final int FOUR_BYTE = 248;//11111000public 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;}