LeetCode赛题393----UTF-8 Validation

393. UTF-8 Validation

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

1. For 1-byte character, the first bit is a 0, followed by its unicode code.
2. 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.

算法分析

算法很简单，只需要依次检查每个数字是否是在合法的范围内即可：如果一个数字在0x00~0x7F之间，说明是 1-byte 字符，检查下一个字符；如果一个数字在0xC0_{0xDF之间，则应为2-byte字符，那么接下来的一个数字应该在0x80}0xBF之间；如果一个数字在0xE0_{0xEF之间，则应为3-byte字符，那么接下来的两个数字应该在0x80}0xBF之间；如果一个数字在0xF0_{0xF7之间，则应为4-byte字符，那么接下来的三个数字应该在0x80}0xBF之间。

Java算法实现：

public class Solution {
    public boolean validUtf8(int[] data) {
        int len=data.length;
        int index=0;
        int num,num1,num2,num3;
        while(index<len){
        	num=data[index];
        	num&=0xff;
        	if(num>=0&&num<=0x7f){
        		//is 1 byte character
        		index++;
        	}
        	else if(num>=0xc0&&num<=0xdf){
        		//is 2-byte character
        		if(index+1<len){
        			num1=data[index+1];
        			num1&=0xff;
        			if(!(num1<=0xbf&&num1>=0x80)){
        				return false;
        			}
        			//the second byte is right
        			index+=2;
        		}
        		else{
        			return false;
        		}
        	}
        	else if(num>=0xe0&&num<=0xef){
        		//it is a 3-byte character
        		if(index+2<len){
        			num1=data[index+1];
        			num2=data[index+2];
        			num1&=0xff;
        			num2&=0xff;
        			if(!(num1>=0x80&&num1<=0xbf&&num2>=0x80&&num2<=0xbf)){
        				return false;
        			}
        			index+=3;
        		}
        		else{
        			return false;
        		}
        	}
        	else if(num>=0xf0&&num<=0xf7){
        		//is a 4-byte character
        		if(index+3<len){
        			num1=data[index+1];
        			num2=data[index+2];
        			num3=data[index+3];
        			num1&=0xff;
        			num2&=0xff;
        			num3&=0xff;
        			if(!(num1>=0x80&&num1<=0xbf&&num2>=0x80&&num2<=0xbf&&num3>=0x80&&num3<=0xbf)){
        				return false;
        			}
        			index+=4;
        		}
        		else{
        			return false;
        		}
        	}
        	else{
        		return false;
        	}
        }
        return true;
    }
}