Java 如何破解谷歌&x27;s多段线算法?
: 你怎么破译这个Java 如何破解谷歌&x27;s多段线算法?,java,google-maps,language-agnostic,Java,Google Maps,Language Agnostic,: 你怎么破译这个 可能是反向运行算法;但是我被困在第5步:如果没有初始值,我怎么知道它是正的还是负的?它的编码左移,如果是负的,则反转,例如: 1: 0000_0001 =>0000_0010 2: 0000_0010 =>0000_0100 3: 0000_0011 =>0000_0110 4: 0000_0100 =>0000_1000 5: 0000_0101 =>0000_1010 6: 0000_0110 =>0000_1100 7: 0000_
可能是反向运行算法;但是我被困在第5步:如果没有初始值,我怎么知道它是正的还是负的?它的编码左移,如果是负的,则反转,例如:
1: 0000_0001 =>0000_0010
2: 0000_0010 =>0000_0100
3: 0000_0011 =>0000_0110
4: 0000_0100 =>0000_1000
5: 0000_0101 =>0000_1010
6: 0000_0110 =>0000_1100
7: 0000_0111 =>0000_1110
8: 0000_1000 =>0001_0000
-1: 1111_1111 =>1111_1110 =>0000_0001
-2: 1111_1110 =>1111_1100 =>0000_0011
-3: 1111_1101 =>1111_1010 =>0000_0101
-4: 1111_1100 =>1111_1000 =>0000_0111
-5: 1111_1011 =>1111_0110 =>0000_1001
-6: 1111_1010 =>1111_0100 =>0000_1011
-7: 1111_1001 =>1111_0010 =>0000_1101
-8: 1111_1000 =>1111_0000 =>0000_1111
因此,如果最后一位是0
,则解码具有正初始值,如果最后一位是1
,则解码具有负初始值<人力资源>附录:
完整解码演示:
public class Test {
public static void main(String args[]) {
for (int point : Decode("_p~iF~ps|U_ulLnnqC_mqNvxq`@",10)) {
System.out.println(point); // Be aware that point is in E5
}
}
private static java.util.List<java.lang.Integer> Decode(String encoded_polylines, int initial_capacity) {
java.util.List<java.lang.Integer> trucks = new java.util.ArrayList<java.lang.Integer>(initial_capacity);
int truck = 0;
int carriage_q = 0;
for (int x = 0, xx = encoded_polylines.length(); x < xx; ++x) {
int i = encoded_polylines.charAt(x);
i -= 63;
int _5_bits = i << (32 - 5) >>> (32 - 5);
truck |= _5_bits << carriage_q;
carriage_q += 5;
boolean is_last = (i & (1 << 5)) == 0;
if (is_last) {
boolean is_negative = (truck & 1) == 1;
truck >>>= 1;
if (is_negative) {
truck = ~truck;
}
trucks.add(truck);
carriage_q = 0;
truck = 0;
}
}
return trucks;
}
}
公共类测试{
公共静态void main(字符串参数[]){
for(int point:Decode(“_p~iF~ps|U ullnqc_mqNvxq`@”,10)){
System.out.println(点);//注意点在E5中
}
}
私有静态java.util.List解码(字符串编码的多段线,int初始容量){
java.util.List trucks=new java.util.ArrayList(初始容量);
int卡车=0;
int carries_q=0;
对于(int x=0,xx=encoded_polylines.length();x>(32-5);
卡车|=_5_位>=1;
如果(是负的){
卡车=~卡车;
}
卡车。添加(卡车);
车厢q=0;
卡车=0;
}
}
返回卡车;
}
}
由于这是一个与语言无关的问题,我将从以下位置添加此PHP解决方案(因为PHP中不存在
运算符):
函数decodePolylineToArray($encoded)
{
$length=strlen($encoded);
$index=0;
$points=array();
$lat=0;
$lng=0;
而($index<$length)
{
$b=0;
$shift=0;
$result=0;
做
{
$b=ord(substr($encoded,$index++)-63;
$result |=($b&0x1f)=0x20);
$dlat=($result&1)~($result>>1):($result>>1));
$lat+=$dlat;
$shift=0;
$result=0;
做
{
$b=ord(substr($encoded,$index++)-63;
$result |=($b&0x1f)=0x20);
$dlng=($result&1)~($result>>1):($result>>1));
$lng+=$dlng;
$points[]=阵列($lat*1e-5,$lng*1e-5);
}
返回$points;
}
更新:
解码指令几乎很简单,为了找到原始值,您可以通过已从ASCII字符转换的每个值的最后一位来计算它是正还是负
例:
步骤5.如果值chunked(chunks of 5)的最后一位为“0x1f”,则该值为负值,应将其反转
例如:
|=($foo&0x1f)虽然这个代码片段可以解决这个问题,但它确实有助于提高您文章的质量。请记住,您是在为将来的读者回答这个问题,而这些人可能不知道您提出代码建议的原因。
function decodePolylineToArray($encoded)
{
$length = strlen($encoded);
$index = 0;
$points = array();
$lat = 0;
$lng = 0;
while ($index < $length)
{
$b = 0;
$shift = 0;
$result = 0;
do
{
$b = ord(substr($encoded, $index++)) - 63;
$result |= ($b & 0x1f) << $shift;
$shift += 5;
}
while ($b >= 0x20);
$dlat = (($result & 1) ? ~($result >> 1) : ($result >> 1));
$lat += $dlat;
$shift = 0;
$result = 0;
do
{
$b = ord(substr($encoded, $index++)) - 63;
$result |= ($b & 0x1f) << $shift;
$shift += 5;
}
while ($b >= 0x20);
$dlng = (($result & 1) ? ~($result >> 1) : ($result >> 1));
$lng += $dlng;
$points[] = array($lat * 1e-5, $lng * 1e-5);
}
return $points;
}