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：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; }
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