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Leetcode: UTF-8 Validation
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.
这道题题干给出了判断 one single UTF-8 char的方法,然后给一个UTF-8 char sequence,判断是不是正确sequence. (读题读了很久)
这道题关键是要学到用 & 取出一个bit sequence当中几位的方法
二进制数表示法:在前面加 0b, 八进制加0o, 十六进制加0x
1 public class Solution { 2 public boolean validUtf8(int[] data) { 3 if (data=http://www.mamicode.com/=null || data.length==0) return false; 4 for (int i=0; i<data.length; i++) { 5 if (data[i] > 255) return false; 6 int moreChecks = 0; //moreCheck is the number of more bytes that need to check for this char 7 if ((data[i] & 0b10000000) == 0) moreChecks = 0; 8 else if ((data[i] & 0b11100000) == 0b11000000) moreChecks = 1; 9 else if ((data[i] & 0b11110000) == 0b11100000) moreChecks = 2; 10 else if ((data[i] & 0b11111000) == 0b11110000) moreChecks = 3; 11 else return false; 12 for (int j=1; j<=moreChecks; j++) { 13 if (i+j >= data.length) return false; 14 if ((data[i+j] & 0b11000000) != 0b10000000) return false; 15 } 16 i = i + moreChecks; 17 } 18 return true; 19 } 20 }
Leetcode: UTF-8 Validation
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