【leetcode】393. UTF-8 Validation

一、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:

   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:

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.

思路：不必将数化成二进制，直接比较大小就可以。

c++代码（16ms，37.30%）

class Solution {public:    bool validUtf8(vector<int>& data) {        int len=data.size();        int cnt=0;   //用来记录后面应该要几个10xxxxxx        int i=0;        while(i<len){            if(cnt > 0){                if(data[i] < 192 && data[i]>=128){                    cnt--;                }else{                    return false;                }            }else if(data[i] >= 248){                return false;            }else if(data[i] >= 240){                cnt = 3;            }else if(data[i] >= 224){                cnt = 2;            }else if(data[i] >= 192){                cnt = 1;            }else if(data[i] >= 128){                return false;            }            i++;        }//while        return cnt==0;    }};