[Leetcode] 393. UTF-8 Validation 解题报告

题目：

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：

首先判断一共有几个合法字节，然后对于每个字节，判断其前两个字节是否是“0b10”，思路挺直观的。

代码：

class Solution {public:    bool validUtf8(vector<int>& data) {        int count = 0;          for(const auto &val: data) {              if(count == 0) {                // check the bytes count                if((val >> 5) == 0b110) {                    count = 1;                }                else if((val >> 4) == 0b1110) {                    count = 2;                  }                else if((val >> 3) == 0b11110) {                     count = 3;                }                else if(val >> 7) {// the most significant bit cannot be 0                    return false;                }             }             else {                if((val >> 6) != 0b10) {    // checking the leading bits in each byte                    return false;                  }                --count;            }        }          return count == 0;    }};