Question

393. UTF-8 Validation

中文文档

Description

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:

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:

   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

Solutions

Solution 1

class Solution:
  def validUtf8(self, data: List[int]) -> bool:
    n = 0
    for v in data:
      if n > 0:
        if v >> 6 != 0b10:
          return False
        n -= 1
      elif v >> 7 == 0:
        n = 0
      elif v >> 5 == 0b110:
        n = 1
      elif v >> 4 == 0b1110:
        n = 2
      elif v >> 3 == 0b11110:
        n = 3
      else:
        return False
    return n == 0

class Solution {
  public boolean validUtf8(int[] data) {
    int n = 0;
    for (int v : data) {
      if (n > 0) {
        if (v >> 6 != 0b10) {
          return false;
        }
        --n;
      } else if (v >> 7 == 0) {
        n = 0;
      } else if (v >> 5 == 0b110) {
        n = 1;
      } else if (v >> 4 == 0b1110) {
        n = 2;
      } else if (v >> 3 == 0b11110) {
        n = 3;
      } else {
        return false;
      }
    }
    return n == 0;
  }
}

class Solution {
public:
  bool validUtf8(vector<int>& data) {
    int n = 0;
    for (int& v : data) {
      if (n > 0) {
        if (v >> 6 != 0b10) return false;
        --n;
      } else if (v >> 7 == 0)
        n = 0;
      else if (v >> 5 == 0b110)
        n = 1;
      else if (v >> 4 == 0b1110)
        n = 2;
      else if (v >> 3 == 0b11110)
        n = 3;
      else
        return false;
    }
    return n == 0;
  }
};

func validUtf8(data []int) bool {
  n := 0
  for _, v := range data {
    if n > 0 {
      if v>>6 != 0b10 {
        return false
      }
      n--
    } else if v>>7 == 0 {
      n = 0
    } else if v>>5 == 0b110 {
      n = 1
    } else if v>>4 == 0b1110 {
      n = 2
    } else if v>>3 == 0b11110 {
      n = 3
    } else {
      return false
    }
  }
  return n == 0
}