393. UTF-8 Validation
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code. 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 bitsof 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.
方法
别人家的代码，自己在convert的部分做了修改，因为255 => 11111111，但是其不满足 11110***，最后剩下的0b10000000使255返回-1：
小技巧:
在java中与操作可以 a & 0b1111111,其中0b表示二进制。
做变量判断时，最好使用–，然后判断是否等于想要值，这样可以避免使用另外变量再去判断，节省了存储空间
public class Solution {
public boolean validUtf8(int[] data) {
int num = 0;
for(int i = 0; i < data.length; i++){
if(num > 0){
if((data[i] & 0b10000000) == 0b10000000){
num--;
}else{
return false;
}
}else{
num = convert(data[i]);
if(num < 0){
return false;
}
}
}
return num == 0;
}
public int convert(int num){
if (num >= 240 && num <= 247) return 3;
if (num >= 224 && num <= 239) return 2;
if (num >= 192 && num <= 223) return 1;
if ((num & 0b10000000) == 0b10000000) return -1;
return 0;
}
}
自己写的代码太冗余了，自己都不想看，得继续好好加油了！！！
public class Solution {
public boolean validUtf8(int[] data) {
String[] num = new String[data.length];
int count = 1;
int c = 1;
int pos = 0;
for(int i = 0; i < data.length; i++ ){
num[i] = convert(data[i]);
if(num[i].substring(0,1).equals("0")){
if(count != c){
return false;
}
pos = 0;
count = 1;
c = 1;
}else if(num[i].substring(0,3).equals("110")){
if(count != c){
return false;
}
pos = 0;
count = 2;
c = 1;
}else if(num[i].substring(0,4).equals("1110")){
if(count != c){
return false;
}
pos = 0;
count = 3;
c = 1;
}else if(num[i].substring(0,5).equals("11110")){
if(count != c){
return false;
}
pos = 0;
count = 4;
c = 1;
}else if(num[i].substring(0,2).equals("10")){
c++;
}else{
return false;
}
if(++pos == count && count != c){
return false;
}
}
if(count != c){
return false;
}
return true;
}
public String convert(int data){
StringBuffer sb = new StringBuffer();
while(data != 0){
sb.append(data%2);
data = data / 2;
}
if(sb.length() != 8){
int dis = 8-sb.length();
for(int i = 1; i <= dis; i++){
sb.append(0);
}
}
return sb.reverse().toString();
}
}