393-UTF-8 Validation

Description

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.

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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.

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問題描述

UTF編碼下的字元可以由1到4個位元組組成， 遵守如下規則:

1. 一個位元組長度的字元， 第一位為0， 接下來的位為unicode編碼
2. n個位元組長度的字元， 開始的n位都是1， 第n + 1位為0， 接下來的n - 1個位元組的最高的兩位為10

給定一個陣列代表資料， 返回資料是否為有效的UTF8編碼。

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問題分析

迭代data, 遍歷每個整數的時候， 首先算出該字元的位元組個數， 然後看看這一個字元對應的這些位是否滿足情況即可。

需要注意一些位運算的技巧

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解法

class Solution {
    public boolean validUtf8(int[] data) {
        if(data == null || data.length == 0) return false;

        boolean isValid = true;
        for(int i = 0;i < data.length;i++){
            if(data[i] > 255) return false; // 1 after 8th digit, 100000000
            int numberOfBytes = 0;
            if((data[i] & 128) == 0){ // 0xxxxxxx, 1 byte, 128(10000000)
                numberOfBytes = 1;
            }else if((data[i] & 224) == 192){ // 110xxxxx, 2 bytes, 224(11100000), 192(11000000)
                numberOfBytes = 2;
            }else if((data[i] & 240) == 224){ // 1110xxxx, 3 bytes, 240(11110000), 224(11100000)
                numberOfBytes = 3;
            }else if((data[i] & 248) == 240){ // 11110xxx, 4 bytes, 248(11111000), 240(11110000)
                numberOfBytes = 4;
            }else{
                //一個字元只由1到4個位元組組成
                return false;
            }
            //判斷該字元的位元組是否滿足規則
            for(int j = 1;j < numberOfBytes;j++){ // check that the next n bytes start with 10xxxxxx
                if(i + j >= data.length) return false;
                if((data[i + j] & 192) != 128) return false; // 192(11000000), 128(10000000)
            }
            //注意這裡
            i = i + numberOfBytes - 1;
        }

        return isValid;
    }
}