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393 UTF-8 Validation

1. Question

A character in UTF8 can be from1 to 4 bytes long, subjected to the following rules:
  1. 1.
    For 1-byte character, the first bit is a 0, followed by its unicode code.
  2. 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:
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Char. number range | UTF-8 octet sequence
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(hexadecimal) | (binary)
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--------------------+---------------------------------------------
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0000 0000-0000 007F | 0xxxxxxx
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0000 0080-0000 07FF | 110xxxxx 10xxxxxx
6
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
7
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
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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:
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data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
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Return true.
4
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
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Example 2:
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data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
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Return false.
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The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
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The next byte is a continuation byte which starts with 10 and that's correct.
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But the second continuation byte does not start with 10, so it is invalid.
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2. Implementation

(1) Bit Manipulation
思路: 这题基本就按照题目给的utf-8的特性去验证,我们用一个变量count检查接下来的有多少个byte需要判断其是否以10开头的。如果count是0的话,我们将data[i]右移相应的位数,比如将其右移3位,判断右移后的数是否等于11110,如果是的话说明是一个4byte的character, 注意如果count是0,而一个byte是10000000的话,这不是valid的utf-8格式
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class Solution {
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public boolean validUtf8(int[] data) {
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int count = 0;
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for (int oneByte : data) {
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if (count == 0) {
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if ((oneByte >> 3) == 0b11110) {
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count = 3;
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}
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else if ((oneByte >> 4) == 0b1110) {
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count = 2;
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}
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else if ((oneByte >> 5) == 0b110) {
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count = 1;
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}
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else if ((oneByte >> 7) == 0b1) {
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return false;
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}
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}
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else {
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if ((oneByte >> 6) == 0b10) {
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--count;
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}
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else {
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return false;
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}
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}
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}
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return count == 0;
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}
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}
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3. Time & Space Complexity

Bit Manipulation: 时间复杂度O(N), 空间复杂度O(1)