【Leetcode】UTF-8 Validation

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最新推荐文章于 2025-07-23 15:25:26 发布

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分类专栏： leetcode Leetcode题解java版

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本文介绍了一种用于验证整数数组是否为有效UTF-8编码的方法。通过解析UTF-8编码规则，实现了一个判断算法，并给出了具体示例。文章详细解释了如何逐字节检查输入数据的有效性。

题目链接：https://leetcode.com/problems/utf-8-validation/

题目：

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.

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.

思路：

easy，不过代码有点丑，尤其那个judge。。判断从前往后有多少个连续1的。。

算法：

	public int judge(int n) {
		if ((128 & n) == 0)
			return -1;
		if ((192 & n) == 128)//10
			return 0;
		if ((224 & n) == 192)
			return 1;
		if ((240 & n) == 224)
			return 2;
		if ((248 & n) == 240)
			return 3;
		return -2;
	}

	public boolean validUtf8(int[] data) {
		boolean isUtf8 = false;// 当前num是否在一个验证的字符中
		int times = 0;
		for (int i = 0; i < data.length; i++) {
			if(isUtf8==false){//当前数不在某字符编码内
				times = judge(data[i]);//当前编码需要多少个10
				if (times == -2 || times == 0)//非编码
					return false;
				if(times>0)//后面需要跟多少个10编码
					isUtf8=true;
			}else{
				if(times>0&&judge(data[i])==0)//是10编码
					times--;
				else
					return false;
				if(times==0)
					isUtf8=false;
			}
		}
		return times>0?false:true;
	}