Leetcode-393. UTF-8 Validation

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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.

说实话，对这种题目很长的，第一感觉就是很蛋疼。这个就是用与操作判断有多少个unicode然后用规则判断即可。 Your runtime beats 36.14% of java submissions.

public class Solution {
    public boolean validUtf8(int[] data) {
        int[] filters = new int[]{128, 192, 224, 240};
        int index = 0;
        boolean flag = true;
        while(index < data.length){
            if((data[index] & 248) == 248) return false;
            if((data[index] & filters[0]) == 0) index ++;
            else{
                flag = false;
                
                for(int i = filters.length - 1; i >0 ; i --){

                    if((data[index] & filters[i]) == filters[i]){
                        flag = true;
                        int count = i;
                        if(data.length - index < count) {flag = false; break;}
                        else{
                            for(int j = index + 1; j <= index + count; j ++){
                                if((data[j] & filters[1]) != filters[0]){flag = false; break;}
                            }
                        }
                        index += count + 1;
                        break;
                    }
                    
                }
                if(!flag) break;
            }
            if(!flag) break;
        }

        return flag;
    }
}