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 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; } }
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