Wednesday, September 7, 2016

LeetCode 393 - UTF-8 Validation


https://leetcode.com/problems/utf-8-validation/
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
  1. For 1-byte character, the first bit is a 0, followed by its unicode code.
  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:
   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
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:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.

Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.

Return false.
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.
http://www.cnblogs.com/grandyang/p/5847597.html
这种方法也是要记连续1的个数,如果是标识字节,先将其向右平移五位,如果得到110,则说明后面跟了一个字节,否则向右平移四位,如果得到1110,则说明后面跟了两个字节,否则向右平移三位,如果得到11110,则说明后面跟了三个字节,否则向右平移七位,如果为1的话,说明是10000000这种情况,不能当标识字节,直接返回false。在非标识字节中,向右平移六位,如果得到的不是10,则说明不是以10开头的,直接返回false,否则cnt自减1,成功完成遍历返回true

    bool validUtf8(vector<int>& data) {
        int cnt = 0;
        for (int d : data) {
            if (cnt == 0) {
                if ((d >> 5) == 0b110) cnt = 1;
                else if ((d >> 4) == 0b1110) cnt = 2;
                else if ((d >> 3) == 0b11110) cnt = 3;
                else if (d >> 7) return false;
            } else {
                if ((d >> 6) != 0b10) return false;
                --cnt;
            }
        }
        return cnt == 0;
    }
http://blog.csdn.net/mebiuw/article/details/52445248
而解题的核心在于: 
1、按照规则,识别byte应该有几位 
2、如果是2-4bytes的数据,继续检查后面的数据是否10开头再返回1,而是1byte则直接返回
public boolean validUtf8(int[] data) { int n = data.length; int skip = 0b10000000; int check = 0; for (int i = 0; i < data.length; i++) { if (check > 0) { if ((data[i] & skip) == skip) check--; else return false; } else { check = getHeadType(data[i]); if (check < 0) return false; } } return check == 0; } /** * 检查*/ public int getHeadType(int num) { if ((num & 0b11110000) == 0b11110000) return 3; if ((num & 0b11100000) == 0b11100000) return 2; if ((num & 0b11000000) == 0b11000000) return 1; if ((num & 0b10000000) == 0b10000000) return -1; //error return 0; }

http://bookshadow.com/weblog/2016/09/04/leetcode-utf-8-validation/
def validUtf8(self, data): """ :type data: List[int] :rtype: bool """ masks = [0x0, 0x80, 0xE0, 0xF0, 0xF8] bits = [0x0, 0x0, 0xC0, 0xE0, 0xF0] while data: for x in (4, 3, 2, 1, 0): if data[0] & masks[x] == bits[x]: break if x == 0 or len(data) < x: return False for y in range(1, x): if data[y] & 0xC0 != 0x80: return False data = data[x:] return True



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