87 lines
2.8 KiB
C++
87 lines
2.8 KiB
C++
// Source : https://leetcode.com/problems/utf-8-validation/
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// Author : Hao Chen
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// Date : 2016-09-08
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/***************************************************************************************
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*
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* A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
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*
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* For 1-byte character, the first bit is a 0, followed by its unicode code.
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* For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by
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* n-1 bytes with most significant 2 bits being 10.
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*
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* This is how the UTF-8 encoding would work:
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*
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* Char. number range | UTF-8 octet sequence
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* --------------------+---------------------------------------------
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* 0000 0000-0000 007F | 0xxxxxxx
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* 0000 0080-0000 07FF | 110xxxxx 10xxxxxx
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* 0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
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* 0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
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*
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* Given an array of integers representing the data, return whether it is a valid utf-8
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* encoding.
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*
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* Note:
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* The input is an array of integers. Only the least significant 8 bits of each integer
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* is used to store the data. This means each integer represents only 1 byte of data.
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*
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* Example 1:
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*
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* data = [197, 130, 1], which represents the octet sequence: 11000101 10000010
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* 00000001.
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*
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* Return true.
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* It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
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*
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* Example 2:
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*
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* data = [235, 140, 4], which represented the octet sequence: 11101011 10001100
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* 00000100.
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*
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* Return false.
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* The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
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* The next byte is a continuation byte which starts with 10 and that's correct.
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* But the second continuation byte does not start with 10, so it is invalid.
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***************************************************************************************/
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class Solution {
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public:
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bool validUtf8(vector<int>& data) {
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int i = 0;
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while ( i < data.size() ) {
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if ( (data[i] & 0x80) == 0 ){
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i++;
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continue;
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}
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int len = 0;
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if ( (data[i] & 0xE0) == 0xC0 ) { // checking 110xxxxx
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len = 2;
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}else if ( (data[i] & 0xF0) == 0xE0) { // checking 1110xxxx
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len = 3;
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}else if ( (data[i] & 0xF8) == 0xF0) { // checking 11110xxx
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len = 4;
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}else {
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return false;
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}
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for (int j=i+1; j < i+len; j++) { //checking 10xxxxxx
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if ( (data[j] & 0xC0) != 0x80 ) {
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return false;
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}
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}
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i += len ;
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if (i > data.size()) {
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return false;
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}
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}
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return true;
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}
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};
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