393. UTF-8 Validation

Description

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.

Problem URL

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Solution

判断一系列的用整数表示的UTF-8字符是否有效。首先判断有几位字符（1-4）位，其他情况非法；然后判断后续字符是否合法。切记判断完后面之后的指针移动。

We could use bit manipulation to resolve this problem. Find the character’s length first. Because of the special UTF-8 encoding sequence, we could use &(and) to decide whether it is valid or not.

Code

class Solution {
    public boolean validUtf8(int[] data) {
        if (data.length == 0)
            return false;
        for (int i = 0; i < data.length; i++){
            if (data[i] >= 255)
                return false;
            int numberOfBytes = 0;
            if ((data[i] & 128) == 0){
                numberOfBytes = 1;
            }
            else if ((data[i] & 224) == 192){ //110x xxxx > 192, 1110 0000 = 224, 1100 0000 = 192
                numberOfBytes = 2;
            }
            else if ((data[i] & 240) == 224){ //1110 xxxx > 224, 1111 0000 = 240, 1110 0000 = 224
                numberOfBytes = 3;
            }
            else if ((data[i] & 248) == 240){ //1111 0xxx > 240, 1111 1000 = 248, 1111 0000 = 240
                numberOfBytes = 4;
            }
            else{
                return false; // other invalid circumstance
            }
            
            for (int j = 1; j < numberOfBytes; j++){
                if (i + j >= data.length)
                    return false;
                if ((data[i+j] & 192) != 128) //10xx xxxx > 12, 1100 0000 = 192, 1000 0000 = 128
                    return false;
            }
            i = i + numberOfBytes - 1;
        }
        return true;
    }
}

Time Complexity: O(n)
Space Complexity: O(1)

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Review

A little block on comprehension of this problem.
I = I + numberOfBytes - 1. It is the issue of I++, we do not want the redundant plus.