LeetCode UTF-8 Validation

LeetCode UTF-8 Validation

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

1. For 1-byte character, the first bit is a 0, followed by its unicode code.
2. 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
--------------------+---------------------------------------------
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.```

```class Solution {
public:
bool validUtf8(vector<int>& data) {
vector<int> mask1 = { 0x80,0xe0,0xf0,0xf8 };
vector<int> first = { 0x0,0xc0,0xe0,0xf0 };
int mask2 = 0xc0, second = 0x80;
int i = 0, j = 0;
while (i < data.size()) {
for (j = 0; j < 4; ++j) {
if ((data[i] & mask1[j]) == first[j]) {
while (j--) {
if (++i >= data.size())return false;
if ((data[i] & mask2) != second)return false;
}
break;
}
}
if (j >= 4)return false;
++i;
}
return true;
}
};
```