LeetCode UTF-8 Validation

LeetCode UTF-8 Validation

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.

给定一个数组,问这个数组是否能表示正确的UTF8字符。UTF8格式在题目中有详细的说明,字节数为1-4个字节,1字节的UTF8第一个bit为0,n字节的UTF8,前n-bits为1,后面跟一个0bit,后面n-1字节前两个bit为10。

解法为我们依次取出字节的前1,3,4,5bits,看是否是1-4字节的UTF8,检查完第一个字节后,根据字节数检查后面的n-1字节。

注意UTF8字节数只可能是1-4,如果发现第一个字节的前5个及以上的bit都是1,直接返回false。代码如下:

class Solution {
public:
	bool validUtf8(vector<int>& data) {
		vector<int> mask1 = { 0x80,0xe0,0xf0,0xf8 };
		vector<int> first = { 0x0,0xc0,0xe0,0xf0 };
		int mask2 = 0xc0, second = 0x80;
		int i = 0, j = 0;
		while (i < data.size()) {
			for (j = 0; j < 4; ++j) {
				if ((data[i] & mask1[j]) == first[j]) {
					while (j--) {
						if (++i >= data.size())return false;
						if ((data[i] & mask2) != second)return false;
					}
					break;
				}
			}
			if (j >= 4)return false;
			++i;
		}
		return true;
	}
};

注意位运算的优先级小于判等运算,所以一定记得加括号,要不然WA。。。

本代码提交AC,用时16MS。

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