前言:为了后续的实习面试,开始疯狂刷题,非常欢迎志同道合的朋友一起交流。因为时间比较紧张,目前的规划是先过一遍,写出能想到的最优算法,第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN,mcf171专栏。
博客链接:mcf171的博客
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A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
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.说实话,对这种题目很长的,第一感觉就是很蛋疼。这个就是用与操作判断有多少个unicode然后用规则判断即可。 Your runtime beats 36.14% of java submissions.
public class Solution {
public boolean validUtf8(int[] data) {
int[] filters = new int[]{128, 192, 224, 240};
int index = 0;
boolean flag = true;
while(index < data.length){
if((data[index] & 248) == 248) return false;
if((data[index] & filters[0]) == 0) index ++;
else{
flag = false;
for(int i = filters.length - 1; i >0 ; i --){
if((data[index] & filters[i]) == filters[i]){
flag = true;
int count = i;
if(data.length - index < count) {flag = false; break;}
else{
for(int j = index + 1; j <= index + count; j ++){
if((data[j] & filters[1]) != filters[0]){flag = false; break;}
}
}
index += count + 1;
break;
}
}
if(!flag) break;
}
if(!flag) break;
}
return flag;
}
}