leetcode393. UTF-8 Validation

题目要求

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.

检验整数数组能否构成合法的UTF8编码的序列。UTF8的字节编码规则如下:

  1. 每个UTF8字符包含1~4个字节
  2. 如果只包含1个字节,则该字节以0作为开头,剩下的位随意
  3. 如果包含两个或两个以上字节,则起始字节以n个1和1个0开头,例如,如果该UTF8字符包含两个字节,则第一个字节以110开头,同理,三个字符的第一个字节以1110开头。剩余的字节必须以10开头。

思路和代码

首先我们整理一下,每一种类型的UTF8字符包含什么样的规格:

  1. 只包含一个字节,该字节格式为0xxxxxxx,则转换为整数的话,该整数必须小于128(1000000)
  2. 包含多个字节,则头字节格式为110xxxxx, 1110xxxx, 11110xxx。而紧跟其后的字符必须格式为10xxxxxx。

综上所述:

  1. num<1000000: 单字节
  2. 10000000=<num<11000000: 多字节字符的跟随字节
  3. 11000000<=num<11100000: 两个字节的起始字节
  4. 11100000<=num<11110000: 三个字节的起始字节
  5. 11110000<=num<11111000: 四个字节的起始字节

下面分别是这题的两种实现:

递归实现:

    private static final int ONE_BYTE = 128; //10000000
    private static final int FOLLOW_BYTE = 192; //11000000
    private static final int TWO_BYTE = 224; //11100000
    private static final int THREE_BYTE = 240;//11110000
    private static final int FOUR_BYTE = 248;//11111000
    public boolean validUtf8(int[] data) {
        return validUtf8(data, 0);
    }
    
    public boolean validUtf8(int[] data, int startAt) {
        if(startAt >= data.length) return true;
        int first = data[startAt];
        
        int followLength = 0;
        if(first < ONE_BYTE) {
            return validUtf8(data, startAt+1);
        }else if(first < FOLLOW_BYTE){
            return false;
        }else if(first <TWO_BYTE) {
            followLength = 2;
        }else if(first < THREE_BYTE) {
            followLength = 3;
        }else if(first < FOUR_BYTE) {
            followLength = 4;
        }else {
            return false;
        }
        if(startAt + followLength > data.length) return false; 
        for(int i = 1 ; i<followLength ; i++) {
            int next = data[startAt + i];
            if(next < ONE_BYTE || next >= FOLLOW_BYTE) {
                return false;
            }
        }
        return validUtf8(data, startAt + followLength);
    }

循环实现:

    private static final int ONE_BYTE = 128; //10000000
    private static final int FOLLOW_BYTE = 192; //11000000
    private static final int TWO_BYTE = 224; //11100000
    private static final int THREE_BYTE = 240;//11110000
    private static final int FOUR_BYTE = 248;//11111000
    public boolean validUtf8(int[] data) {
        return validUtf8(data, 0);
    }
    
    public boolean validUtf8(int[] data, int startAt) {
        int followCount = 0;
        for(int num : data) {
            if(num < ONE_BYTE) {
                if(followCount != 0) {
                    return false;
                }
            }else if(num < FOLLOW_BYTE) {
                if(followCount == 0) {
                    return false;
                }
                followCount--;
            }else if(num < TWO_BYTE) {
                if(followCount != 0) {
                    return false;
                }
                followCount = 1;
            }else if(num < THREE_BYTE) {
                if(followCount != 0) {
                    return false;
                }
                followCount = 2;
            }else if(num < FOUR_BYTE) {
                if(followCount != 0) {
                    return false;
                }
                followCount = 3;
            }else {
                return false;
            }
        }
        return followCount == 0;
    }
    原文作者:后端开发
    原文地址: https://segmentfault.com/a/1190000018180791
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