comments | difficulty | edit_url | tags | ||
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true |
Medium |
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Given an integer array data
representing the data, return whether it is a valid UTF-8 encoding (i.e. it translates to a sequence of valid UTF-8 encoded characters).
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 is0
, followed byn - 1
bytes with the most significant2
bits being10
.
This is how the UTF-8 encoding would work:
Number of Bytes | UTF-8 Octet Sequence | (binary) --------------------+----------------------------------------- 1 | 0xxxxxxx 2 | 110xxxxx 10xxxxxx 3 | 1110xxxx 10xxxxxx 10xxxxxx 4 | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
x
denotes a bit in the binary form of a byte that may be either 0
or 1
.
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:
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.
Example 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.
Constraints:
1 <= data.length <= 2 * 104
0 <= data[i] <= 255
We use a variable
For each integer
- If
$cnt > 0$ , then check if$v$ starts with$10$ . If not, returnfalse
, otherwise decrement$cnt$ . - If the highest bit of
$v$ is$0$ , then$cnt = 0$ . - If the highest two bits of
$v$ are$110$ , then$cnt = 1$ . - If the highest three bits of
$v$ are$1110$ , then$cnt = 2$ . - If the highest four bits of
$v$ are$11110$ , then$cnt = 3$ . - Otherwise, return
false
.
Finally, if true
, otherwise return false
.
The time complexity is data
. The space complexity is
class Solution:
def validUtf8(self, data: List[int]) -> bool:
cnt = 0
for v in data:
if cnt > 0:
if v >> 6 != 0b10:
return False
cnt -= 1
elif v >> 7 == 0:
cnt = 0
elif v >> 5 == 0b110:
cnt = 1
elif v >> 4 == 0b1110:
cnt = 2
elif v >> 3 == 0b11110:
cnt = 3
else:
return False
return cnt == 0
class Solution {
public boolean validUtf8(int[] data) {
int cnt = 0;
for (int v : data) {
if (cnt > 0) {
if (v >> 6 != 0b10) {
return false;
}
--cnt;
} else if (v >> 7 == 0) {
cnt = 0;
} else if (v >> 5 == 0b110) {
cnt = 1;
} else if (v >> 4 == 0b1110) {
cnt = 2;
} else if (v >> 3 == 0b11110) {
cnt = 3;
} else {
return false;
}
}
return cnt == 0;
}
}
class Solution {
public:
bool validUtf8(vector<int>& data) {
int cnt = 0;
for (int& v : data) {
if (cnt > 0) {
if (v >> 6 != 0b10) {
return false;
}
--cnt;
} else if (v >> 7 == 0) {
cnt = 0;
} else if (v >> 5 == 0b110) {
cnt = 1;
} else if (v >> 4 == 0b1110) {
cnt = 2;
} else if (v >> 3 == 0b11110) {
cnt = 3;
} else {
return false;
}
}
return cnt == 0;
}
};
func validUtf8(data []int) bool {
cnt := 0
for _, v := range data {
if cnt > 0 {
if v>>6 != 0b10 {
return false
}
cnt--
} else if v>>7 == 0 {
cnt = 0
} else if v>>5 == 0b110 {
cnt = 1
} else if v>>4 == 0b1110 {
cnt = 2
} else if v>>3 == 0b11110 {
cnt = 3
} else {
return false
}
}
return cnt == 0
}
function validUtf8(data: number[]): boolean {
let cnt = 0;
for (const v of data) {
if (cnt > 0) {
if (v >> 6 !== 0b10) {
return false;
}
--cnt;
} else if (v >> 7 === 0) {
cnt = 0;
} else if (v >> 5 === 0b110) {
cnt = 1;
} else if (v >> 4 === 0b1110) {
cnt = 2;
} else if (v >> 3 === 0b11110) {
cnt = 3;
} else {
return false;
}
}
return cnt === 0;
}