题目描述
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:
- For a 1-byte character, the first bit is a
0
, followed by its Unicode code.- For an n-bytes character, the first
n
bits are all one’s, then + 1
bit is 0, followed byn - 1
bytes with the most significant2
bits being10
.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 with10
and that’s correct. But the second continuation byte does not start with10
, 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:
#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