Problem

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 (hexadecimal) UTF-8 octet sequence (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

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.

Solution

class Solution {
    public boolean validUtf8(int[] data) {
        if (data == null || data.length == 0) return false;
        for (int i = 0; i < data.length; i++) {
            if (data[i] > 255) return false;
            int count = 0;
            if (data[i] < 128) {
                count = 1;
            } else if (data[i] >= 192 && data[i] < 224) {
                count = 2;
            } else if (data[i] < 240) {
                count = 3;
            } else if (data[i] < 248) {
                count = 4;
            } else {
                return false;
            }
            
            for (int j = 1; j < count; j++) {
                if (i+j >= data.length) return false;
                if (data[i+j] < 128 || data[i+j] >= 192) return false;
            }
            
            i = i+count-1;
        }
        return true;
    }
}

linspiration
161 声望53 粉丝