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86 lines (77 loc) · 2.82 KB
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// Source : https://leetcode.com/problems/utf-8-validation/
// Author : Hao Chen
// Date : 2016-09-08
/***************************************************************************************
*
* 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
* --------------------+---------------------------------------------
* 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) {
int i = 0;
while ( i < data.size() ) {
if ( (data[i] & 0x80) == 0 ){
i++;
continue;
}
int len = 0;
if ( (data[i] & 0xE0) == 0xC0 ) { // checking 110xxxxx
len = 2;
}else if ( (data[i] & 0xF0) == 0xE0) { // checking 1110xxxx
len = 3;
}else if ( (data[i] & 0xF8) == 0xF0) { // checking 11110xxx
len = 4;
}else {
return false;
}
for (int j=i+1; j < i+len; j++) { //checking 10xxxxxx
if ( (data[j] & 0xC0) != 0x80 ) {
return false;
}
}
i += len ;
if (i > data.size()) {
return false;
}
}
return true;
}
};