[刷题防痴呆] 0393 - UTF-8 编码验证 (UTF-8

题目地址

https://leetcode.com/problems/utf-8-validation/

题目描述

393. UTF-8 Validation

Given an integer array data representing the data, return whether it is a valid UTF-8 encoding.

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

For a 1-byte character, the first bit is a 0, followed by its Unicode code.
For an n-bytes character, the first n bits are all one's, the n + 1 bit is 0, followed by n - 1 bytes with the 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
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:

Input: data = [197,130,1]
Output: true
Explanation: data represents the octet sequence: 11000101 10000010 00000001.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:

Input: data = [235,140,4]
Output: false
Explanation: data represented the octet sequence: 11101011 10001100 00000100.
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

• 模拟. 先Integer.toBinaryString, 取最后8位.
• 看总共有多少个需要process的.
• 之后, 按照要求一个个parse比较.

关键点

代码

• 语言支持：Java

class Solution {
    public boolean validUtf8(int[] data) {
        int process = 0;
        for (int i = 0; i < data.length; i++) {
            String str = Integer.toBinaryString(data[i]);
            if (str.length() >= 8) {
                str = str.substring(str.length() - 8, str.length());
            } else {
                str = "00000000".substring(str.length() % 8) + str;
            }

            if (process == 0) {
                for (int j = 0; j < str.length(); j++) {
                    if (str.charAt(j) == '0') {
                        break;
                    }
                    process++;
                }

                if (process == 0) {
                    continue;
                }

                 if (process > 4 || process == 1) {
                    return false;
                }               
            } else {
                if (!(str.charAt(0) == '1' && str.charAt(1) == '0')) {
                    return false;
                }
            }
            process--;
        }

        return process == 0;
    }
}