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utf-8Validation.java
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// 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 Solution {
public boolean validUtf8(int[] data) {
int count = 0;
for(int i : data) {
if(count == 0) {
if((i >> 5) == 0b110) count = 1;
else if((i >> 4) == 0b1110) count = 2;
else if((i >> 3) == 0b11110) count = 3;
else if((i >> 7) == 0b1) return false;
}
else {
if((i >> 6) != 0b10) return false;
count--;
}
}
return count == 0;
}
}