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[Leetcode]393. UTF-8 Validation(C++)

题目描述

题目链接: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:

  1. For a 1-byte character, the first bit is a 0, followed by its Unicode code.
  2. 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

例子

例子 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.

例子 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.

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.

解题思路

这道题理解了题意之后并不难,只需要按照规则一个个推算即可。从第一个字节中通过不同数字相与获得连续字节个数 count 再依次判断接下来 count - 1 个数字是否为 10 开头即可。只要其中有一项不符合直接返回 false

代码如下:

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#include <vector>

class Solution {
public:
bool validUtf8(std::vector<int>& data) {
int idx = 0;
int count = 1;
while (idx < data.size()) {
int first_byte = data[idx];
if ((first_byte & 0x80) == 0x00) {
// use 1000 0000 as mask to see if start with 0
count = 1;
} else if ((first_byte & 0xe0) == 0xc0) {
// use 1110 0000 as mask to see if start with 110
count = 2;
} else if ((first_byte & 0xf0) == 0xe0) {
// use 1111 0000 as mask to see if start with 1110
count = 3;
} else if ((first_byte & 0xf8) == 0xf0) {
// use 1111 1000 as mask to see if start with 11110
count = 4;
} else {
return false;
}

while (count > 1) {
idx++;
count--;
// thr following count - 1 byte should all start with 10
if (idx == data.size() || (data[idx] & 0xc0) != 0x80)
return false;
}
idx++;
}

return count == 1;
}
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(1)

GitHub 代码同步地址: 393.Utf8Validation.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions