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Does anybody have ann taylor coupon code?
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Does anybody have ann taylor coupon code?# Fashion - 美丽时尚
s*3
1
UTF-8 encoding scheme is described below:
0XXXXXXX = this is the entire rune
10XXXXXX = this is a continuation of the rune from the previous byte
110XXXXX = this is the start of a 2-byte rune.
1110XXXX = this is the start of a 3-byte rune.
11110XXX = this is the start of a 4-byte rune.
111110XX = this is the start of a 5-byte rune.
1111110X = this is the start of a 6-byte rune.
11111110 = this is the start of a 7-byte rune.
11111111 = this is the start of a 8-byte rune.
For example, a 3-byte rune would be 1110XXXX 10XXXXXX 10XXXXXX.
Write a function that decides whether a given byte array (or string) is
valid UTF-8 encoded text.
求讨论求指导
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S*y
2
站内联系。虽然journal IF很烂(IF<0.5),还是希望能认真一下,所以请提供正式(学
校或研究所)email以便向编辑推荐。另请附一两行专业背景。
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h*m
3
Please email me if you have. Thanks a lot.
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x*j
4
没太看懂
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S*y
5
Gone.

【在 S********y 的大作中提到】
: 站内联系。虽然journal IF很烂(IF<0.5),还是希望能认真一下,所以请提供正式(学
: 校或研究所)email以便向编辑推荐。另请附一两行专业背景。

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g*l
6
试试welcome1, 25%off full-price
字母要大写

【在 h****m 的大作中提到】
: Please email me if you have. Thanks a lot.
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l*a
7
那就去学习一下UTF-8的定义

【在 x***j 的大作中提到】
: 没太看懂
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l*a
8
通过这个识别每个byte的type,再check是否满足规则?
public int getType(byte input) {
int target=1<<7;
int type=0;
while(target>0) {
if(intput & target==0) return type;
type++;
target>=1;
}
}

【在 s*******3 的大作中提到】
: UTF-8 encoding scheme is described below:
: 0XXXXXXX = this is the entire rune
: 10XXXXXX = this is a continuation of the rune from the previous byte
: 110XXXXX = this is the start of a 2-byte rune.
: 1110XXXX = this is the start of a 3-byte rune.
: 11110XXX = this is the start of a 4-byte rune.
: 111110XX = this is the start of a 5-byte rune.
: 1111110X = this is the start of a 6-byte rune.
: 11111110 = this is the start of a 7-byte rune.
: 11111111 = this is the start of a 8-byte rune.

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l*a
9
Nice~
public boolean validUTF8(byte [] array){
if (array == null || array.length == 0){
return false;
}
int leftByte = 0;
for (int i = 0; i < array.length; i++){
int val = getType(array[i]);
if (leftByte == 0){ // new start
if (val == 1){ // continue byte
return false;
}
else if (val == 0){
continue;
}
else {
leftByte = val-1;
}
}
else{ // shall continue
if (val != 1){
return false;
}
else{
leftByte--;
}
}
}
if (leftByte == 0){
return true;
}
else{
return false;
}
}

public int getType(byte input) {
int target=1<<7;
int type=0;
while(target>0) {
if((input & target)==0) return type;
type++;
target = target>>1;
}
return type;
}

【在 l*****a 的大作中提到】
: 通过这个识别每个byte的type,再check是否满足规则?
: public int getType(byte input) {
: int target=1<<7;
: int type=0;
: while(target>0) {
: if(intput & target==0) return type;
: type++;
: target>=1;
: }
: }

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p*p
10
这样做不是最优吧。没必要把8位都循环一遍。
遇到第一个0就可以返回了:
private int getType(byte b) {
int mask = 1 << 7;
int type = 0;
while((mask & b) != 0) {
type++;
mask >>= 1;
}
return type;
}

【在 l*****a 的大作中提到】
: 通过这个识别每个byte的type,再check是否满足规则?
: public int getType(byte input) {
: int target=1<<7;
: int type=0;
: while(target>0) {
: if(intput & target==0) return type;
: type++;
: target>=1;
: }
: }

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p*p
11
看错了。你的方法也是第一个0就返回。

【在 p**p 的大作中提到】
: 这样做不是最优吧。没必要把8位都循环一遍。
: 遇到第一个0就可以返回了:
: private int getType(byte b) {
: int mask = 1 << 7;
: int type = 0;
: while((mask & b) != 0) {
: type++;
: mask >>= 1;
: }
: return type;

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p*2
12
我回家写一下

【在 l*****a 的大作中提到】
: 通过这个识别每个byte的type,再check是否满足规则?
: public int getType(byte input) {
: int target=1<<7;
: int type=0;
: while(target>0) {
: if(intput & target==0) return type;
: type++;
: target>=1;
: }
: }

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p*p
13
private int getType(byte b) {
int mask = 1 << 7;
int type = 0;
while((mask & b) != 0) {
type++;
mask >>= 1;
}
return type;
}

public boolean isValidUTF8(byte[] bytes) {
if(bytes == null || bytes.length == 0) {
return false;
}
int bytesLeft = 0;
for(byte b : bytes) {
int type = getType(b);
if(type == 0) {
if(bytesLeft != 0) {
return false;
}
continue;
} else if (type == 1) {
if(bytesLeft == 0) {
return false;
}
bytesLeft--;
} else {
if(bytesLeft != 0) {
return false;
}
bytesLeft = type-1;
}
}
return bytesLeft == 0;
}

【在 s*******3 的大作中提到】
: UTF-8 encoding scheme is described below:
: 0XXXXXXX = this is the entire rune
: 10XXXXXX = this is a continuation of the rune from the previous byte
: 110XXXXX = this is the start of a 2-byte rune.
: 1110XXXX = this is the start of a 3-byte rune.
: 11110XXX = this is the start of a 4-byte rune.
: 111110XX = this is the start of a 5-byte rune.
: 1111110X = this is the start of a 6-byte rune.
: 11111110 = this is the start of a 7-byte rune.
: 11111111 = this is the start of a 8-byte rune.

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n*p
14
while(s != NULL && *s != NULL)
{
if (*s & 0x80 == 0) {s++; continue;}
if (*s & 0xC0 == 0x80) return false;
char* t = s+1;
for(int bitmask = 0x40;
bitmask >0 && (*s & bitmask != 0);
bitmask /= 2, t++)
if (*t == NULL || (*t & 0xC0 != 0x80)) return false;
s = t + 1;
}
return true;

【在 s*******3 的大作中提到】
: UTF-8 encoding scheme is described below:
: 0XXXXXXX = this is the entire rune
: 10XXXXXX = this is a continuation of the rune from the previous byte
: 110XXXXX = this is the start of a 2-byte rune.
: 1110XXXX = this is the start of a 3-byte rune.
: 11110XXX = this is the start of a 4-byte rune.
: 111110XX = this is the start of a 5-byte rune.
: 1111110X = this is the start of a 6-byte rune.
: 11111110 = this is the start of a 7-byte rune.
: 11111111 = this is the start of a 8-byte rune.

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p*2
15
def isUTF(xs: List[String]):Boolean = {
xs match {
case Nil => true
case head::tail => head.toList match {
case '0'::_ => isUTF(xs.tail)
case '1'::'0'::_ => false
case _ =>
val len = (x: String) => x.prefixLength(_ == '1')
val l = len(head)
val rest = tail.take(l-1)
rest.size == l-1 && rest.forall(len(_)==1) && isUTF(xs.drop(l))
}
}
}
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l*8
16
bool isUTF8(const string & str) {
int ct = 0;
for (char ch : str) {
int type = 8;
for (ch = ~ch; ch; ch >>= 1)
--type;
if (ct > 0) {
if (type != 1) return false;
--ct;
} else {
if (type == 1) return false;
ct = max(0, type - 1);
}
}
return ct == 0;
}
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