blocks|key|2905246|text|它的编码左移，如果是负数，则反转，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2905247|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|code-block|syntax|javascript|2905248|因此，如果最后一位是0，初始是正的，并且如果最后一位是1，则初始是负的，则解码具有。|offset|length|style|CODE|2905249|2905250|附录：|BOLD|2905251|​|2905252|完整解码demo：|2905253|public+class+Test+{
+public+static+void+main(String+args[])+{
++for+(int+point+:+Decode("_p~iF~ps%7CU_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;+%2B%2Bx)+{
+++int+i+=+encoded_polylines.charAt(x);
+++i+-=+63;
+++int+_5_bits+=+i+<<+(32+-+5)+>>>+(32+-+5);
+++truck+%7C=+_5_bits+<<+carriage_q;
+++carriage_q+%2B=+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;
+}
}|2905254|entityMap^0|0|0|A|1|R|1|0|0|0|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|16|8|@$I|17|J|18|K|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1A|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|X|$]]

Its encoding leftshifts then inverts if negative, eg: 

<pre><code>1: 0000_0001 =&gt;0000_0010
2: 0000_0010 =&gt;0000_0100
3: 0000_0011 =&gt;0000_0110
4: 0000_0100 =&gt;0000_1000
5: 0000_0101 =&gt;0000_1010
6: 0000_0110 =&gt;0000_1100
7: 0000_0111 =&gt;0000_1110
8: 0000_1000 =&gt;0001_0000

-1: 1111_1111 =&gt;1111_1110 =&gt;0000_0001
-2: 1111_1110 =&gt;1111_1100 =&gt;0000_0011
-3: 1111_1101 =&gt;1111_1010 =&gt;0000_0101
-4: 1111_1100 =&gt;1111_1000 =&gt;0000_0111
-5: 1111_1011 =&gt;1111_0110 =&gt;0000_1001
-6: 1111_1010 =&gt;1111_0100 =&gt;0000_1011
-7: 1111_1001 =&gt;1111_0010 =&gt;0000_1101
-8: 1111_1000 =&gt;1111_0000 =&gt;0000_1111
</code></pre>

Thus, decoding has if last bit is <code>0</code>, initial is positive, and if last bit is <code>1</code>, initial is negative. <hr>Appendix:

Full decoding demo:

<pre><code>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&lt;java.lang.Integer&gt; Decode(String encoded_polylines, int initial_capacity) {
 java.util.List&lt;java.lang.Integer&gt; trucks = new java.util.ArrayList&lt;java.lang.Integer&gt;(initial_capacity);
 int truck = 0;
 int carriage_q = 0;
 for (int x = 0, xx = encoded_polylines.length(); x &lt; xx; ++x) {
 int i = encoded_polylines.charAt(x);
 i -= 63;
 int _5_bits = i &lt;&lt; (32 - 5) &gt;&gt;&gt; (32 - 5);
 truck |= _5_bits &lt;&lt; carriage_q;
 carriage_q += 5;
 boolean is_last = (i &amp; (1 &lt;&lt; 5)) == 0;
 if (is_last) {
 boolean is_negative = (truck &amp; 1) == 1;
 truck &gt;&gt;&gt;= 1;
 if (is_negative) {
 truck = ~truck;
 }
 trucks.add(truck);
 carriage_q = 0;
 truck = 0;
 }
 }
 return trucks;
 }
}
</code></pre>

blocks|key|238162|text|因为这是一个与语言无关的问题，所以我将从Peter+Chng's+unitstep+blog添加这个PHP解决方案(因为PHP中不存在>>>操作符)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|238163|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%2B%2B))+-+63;
++++++$result+%7C=+($b+&+0x1f)+<<+$shift;
++++++$shift+%2B=+5;
++++}
++++while+($b+>=+0x20);
++++$dlat+=+(($result+&+1)+?+~($result+>>+1)+:+($result+>>+1));
++++$lat+%2B=+$dlat;

++++$shift+=+0;
++++$result+=+0;
++++do
++++{
++++++$b+=+ord(substr($encoded,+$index%2B%2B))+-+63;
++++++$result+%7C=+($b+&+0x1f)+<<+$shift;
++++++$shift+%2B=+5;
++++}
++++while+($b+>=+0x20);

++++$dlng+=+(($result+&+1)+?+~($result+>>+1)+:+($result+>>+1));
++++$lng+%2B=+$dlng;

++++$points[]+=+array($lat+*+1e-5,+$lng+*+1e-5);
++}
++return+$points;
}|code-block|syntax|javascript|238164|Additional+instructions+from+Google+developers|238165|更新:解码指令几乎很简单，为了找到原始值，你可以根据你已经从ASCII字符转换的每个值的最后一位来计算它是正数还是负数。|238166|例如：|238167|步骤5.如果您的值分块(chunks+of+5)的最后一位是'0x1f‘，那么它是负数，您应该将其反转，如下所示：%7C=+($foo+&+0x1f)+<<+$shift;|238168|00000010+00100101+01000011+11100001|BOLD|238169|4将二进制值右移一位：|238170|11111101+11011010+10111100+00011110|238171|3将二进制值转换为十进制值，记住如果你意识到这是一个负数，那么你必须将它从二的补码Two's+complement中转换出来:如果它是正的，那么就像往常一样转换二进制|238172|11111110+11101101+01011110+00001111|238173|11111110+11101101+01011110+00001110|238174|00000001+00010010+10100001+11110001+<--我们的原始值无符号|238175|2将十进制值乘以1e5，因此将其除以得到初始值：179.98321|238176|1将原始值加上它的符号(如果需要)+-179.98321+(有点数据丢失，但这是非常不相关的)|238177|entityMap|0|LINK|mutability|MUTABLE|url|http://unitstep.net/blog/2008/08/02/decoding-google-maps-encoded-polylines-using-php/|1|https://developers.google.com/maps/documentation/utilities/polylinealgorithm|2|https://en.wikipedia.org/wiki/Two's_complement^0|1V|3|K|Q|0|0|0|0|1A|1|0|0|0|1L|R|0|0|Z|0|0|0|Z|0|15|G|2|0|0|Z|0|0|Z|0|0|Z|0|O|9|0|I|A|0^^$0|@$1|2|3|4|5|6|7|1N|8|@$9|1O|A|1P|B|C]]|D|@$9|1Q|A|1R|1|1S]]|E|$]]|$1|F|3|G|5|H|7|1T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1U|8|@]|D|@$9|1V|A|1W|1|1X]]|E|$]]|$1|M|3|N|5|6|7|1Y|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|1Z|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|20|8|@$9|21|A|22|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|23|8|@$9|24|A|25|B|U]]|D|@]|E|$]]|$1|V|3|W|5|6|7|26|8|@]|D|@]|E|$]]|$1|X|3|Y|5|6|7|27|8|@$9|28|A|29|B|U]]|D|@]|E|$]]|$1|Z|3|10|5|6|7|2A|8|@]|D|@$9|2B|A|2C|1|2D]]|E|$]]|$1|11|3|12|5|6|7|2E|8|@$9|2F|A|2G|B|U]]|D|@]|E|$]]|$1|13|3|14|5|6|7|2H|8|@$9|2I|A|2J|B|U]]|D|@]|E|$]]|$1|15|3|16|5|6|7|2K|8|@$9|2L|A|2M|B|U]]|D|@]|E|$]]|$1|17|3|18|5|6|7|2N|8|@$9|2O|A|2P|B|U]]|D|@]|E|$]]|$1|19|3|1A|5|6|7|2Q|8|@$9|2R|A|2S|B|U]]|D|@]|E|$]]|$1|1B|3|-4|5|6|7|2T|8|@]|D|@]|E|$]]]|1C|$1D|$5|1E|1F|1G|E|$1H|1I]]|1J|$5|1E|1F|1G|E|$1H|1K]]|1L|$5|1E|1F|1G|E|$1H|1M]]]]

Since this is a language-agnostic question, I will add this PHP solution (since the <code>&gt;&gt;&gt;</code> operator does not exist in PHP) from <a href="http://unitstep.net/blog/2008/08/02/decoding-google-maps-encoded-polylines-using-php/" rel="nofollow">Peter Chng's unitstep blog</a>:

<pre><code>function decodePolylineToArray($encoded)
{
 $length = strlen($encoded);
 $index = 0;
 $points = array();
 $lat = 0;
 $lng = 0;

 while ($index &lt; $length)
 {
 $b = 0;
 $shift = 0;
 $result = 0;
 do
 {
 $b = ord(substr($encoded, $index++)) - 63;
 $result |= ($b &amp; 0x1f) &lt;&lt; $shift;
 $shift += 5;
 }
 while ($b &gt;= 0x20);
 $dlat = (($result &amp; 1) ? ~($result &gt;&gt; 1) : ($result &gt;&gt; 1));
 $lat += $dlat;

 $shift = 0;
 $result = 0;
 do
 {
 $b = ord(substr($encoded, $index++)) - 63;
 $result |= ($b &amp; 0x1f) &lt;&lt; $shift;
 $shift += 5;
 }
 while ($b &gt;= 0x20);

 $dlng = (($result &amp; 1) ? ~($result &gt;&gt; 1) : ($result &gt;&gt; 1));
 $lng += $dlng;

 $points[] = array($lat * 1e-5, $lng * 1e-5);
 }
 return $points;
}
</code></pre>

<a href="https://developers.google.com/maps/documentation/utilities/polylinealgorithm" rel="nofollow">Additional instructions from Google developers</a>

Update:
The decoding instructions are almost straightforward, in order to find the original value, you can calculate if it is positive or negative by the last bit on each value you already transform from your ASCII characters.

ex:

Step 5. If your value chunked (chunks of 5) has a last bit '0x1f' then it is negative and you should invert it
as in: <code>|= ($foo &amp; 0x1f) &lt;&lt; $shift;</code>

00000010 00100101 01000011 11100001 

4 Right-shift the binary value one bit:

11111101 11011010 10111100 00011110

3 Convert the binary value to decimal, Remember if you realize this was a negative number, then you must transform it from two's complement,<a href="https://en.wikipedia.org/wiki/Two&#39;s_complement" rel="nofollow">Two's complement</a>: if it was positive then just convert the binary as usual:

11111110 11101101 01011110 00001111

11111110 11101101 01011110 00001110

00000001 00010010 10100001 11110001 &lt;-- our original value unsigned

2 The decimal value was multiplied by 1e5, so divided it to get the initial value:
179.98321

1 Add the original value its sign (if it needs it)
-179.98321
(the is a little bit of data loss, but its quite irrelevant)

<a href="http://code.google.com/apis/maps/documentation/utilities/polylinealgorithm.html" rel="nofollow noreferrer">Google's Encoded Polyline Algorithm Format</a>:

<blockquote>
 <a href="https://i.stack.imgur.com/kkWPg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kkWPg.png" alt="enter image description here"></a>
</blockquote>

How do you decode this?

Perhaps run the algorithm backwards; But I'm stuck at step 5: without the initial value, how do I know if it's positive/negative?

How to decode Google's Polyline Algorithm?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

你是如何解码的？也许向后运行算法；但我被困在步骤5:没有初始值，我如何知道它是正/负？

问如何解码谷歌的折线算法？
EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何解码谷歌的折线算法？EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何解码谷歌的折线算法？
EN