题目描述:

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:

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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:

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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:

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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.

这道题使用位运算就可以了, 前几天刚刚做完18600的第一个datalab, 位运算还是挺easy的.

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class Solution {
public:
    bool validUtf8(vector<int>& data) {
        int i = 0;
        while(i < data.size()){
            if(data[i] & 0x80){
                int oneLen = 0;
                while((0x80 >> oneLen) & data[i]){
                    oneLen++;
                }
                if(oneLen <= 1 || oneLen > 4) return false;
                int j;
                for(j = 1; j < oneLen; j++){
                    if((data[i + j] & 0xc0) != 0x80) return false;
                }
                i = i + j;
            }
            else{
                i++;
            }
        }
        return true;
    }
};