UTF8Validation393.java

/**
 * 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.
 */


public class UTF8Validation393 {
    public boolean validUtf8(int[] data) {
        int L = data.length;
        if (L == 0) return false;

        for (int k = 0; k<L; ) {
            int num = data[k];
            if (getBit(num, 7) == 0) {
                k++;
            } else {
                int n = 1;
                while (getBit(num, (7-n)) == 1) n++;
                if (n == 1 || n > 4 || k+n > L) return false;

                for (int i = 1; i<n; i++) {
                    int t = data[k+i];
                    if (getBit(t, 7) != 1 || getBit(t, 6) != 0) return false;
                }

                k += n;
            }
        }

        return true;
    }


    private int getBit(int num, int pos) {
        return (num >> pos) & 1;
    }


    /**
     * https://discuss.leetcode.com/topic/58338/bit-manipulation-java-6ms
     */
    public boolean validUtf82(int[] data) {
        if(data==null || data.length==0) return false;
        boolean isValid = true;
        for(int i=0;i<data.length;i++) {
            if(data[i]>255) return false; // 1 after 8th digit, 100000000
            int numberOfBytes = 0;
            if((data[i] & 128) == 0) { // 0xxxxxxx, 1 byte, 128(10000000)
                numberOfBytes = 1;
            } else if((data[i] & 224) == 192) { // 110xxxxx, 2 bytes, 224(11100000), 192(11000000)
                numberOfBytes = 2;
            } else if((data[i] & 240) == 224) { // 1110xxxx, 3 bytes, 240(11110000), 224(11100000)
                numberOfBytes = 3;
            } else if((data[i] & 248) == 240) { // 11110xxx, 4 bytes, 248(11111000), 240(11110000)
                numberOfBytes = 4;
            } else {
                return false;
            }
            for(int j=1;j<numberOfBytes;j++) { // check that the next n bytes start with 10xxxxxx
                if(i+j>=data.length) return false;
                if((data[i+j] & 192) != 128) return false; // 192(11000000), 128(10000000)
            }
            i=i+numberOfBytes-1;
        }
        return isValid;
    }

}