1 解題思想

這道題主要是要求判別給定的二進位制序列是否是合法的UTF-8編碼
精簡的說，UTF-8編碼的特性有：
1、長度可以從1byte到4bytes
2、對於1byte的資料，其位元位的最高位是0
3、對於長度2-4的資料，其首個byte的高n-1位全部為1，隨後的n-1個byte，其高2位全部為10

在這道題中，以int的方式標示一個byte，取他的後8位就可以

而解題的核心在於：
1、按照規則，識別byte應該有幾位
2、如果是2-4bytes的資料，繼續檢查後面的資料是否10開頭再返回1，而是1byte則直接返回

按照上面那個過程解決就OK @MebiuW

2 原題

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
      (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.

3 AC解

public class Solution {
    public boolean validUtf8(int[] data) {
        int n = data.length;
        int skip = 0b10000000;
        int check = 0;
        for (int i = 0; i < data.length; i++) {
            if (check > 0) {
                if ((data[i] & skip) == skip) 
                    check--;
                else 
                    return false;
            } else {
                check = getHeadType(data[i]);
                if (check < 0) return false;
            }
        }
        return check == 0;
    }
    /**
     * 檢查*/
    public int getHeadType(int num) {
        if ((num & 0b11110000) == 0b11110000) return 3;
        if ((num & 0b11100000) == 0b11100000) return 2;
        if ((num & 0b11000000) == 0b11000000) return 1;
        if ((num & 0b10000000) == 0b10000000) return -1; //error
        return 0;
    }
}

因為題目簡單，我覺得沒有太多價值，直接參照discuss一個內容寫的寫的