blocks|key|371909|text|一种“便宜”的方法是有第三个数组，其中包含0到n表示其他数组的索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|371910|但是，如果将它们分组在一个类中，那么所遇到的问题就会消失，这两个信息在逻辑上似乎是相关的。然后实现可比较的：https://stackoverflow.com/a/18896422|offset|length|371911|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/18896422^0|0|1I|10|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

A "cheap" way would be to have a third array which would contain 0 to n representing the index of the other arrays

But the problem you are having would disappear if they were grouped in a class, both information seem logically related.
Then you would implement Comparable: <a href="https://stackoverflow.com/a/18896422">https://stackoverflow.com/a/18896422</a>

blocks|key|371926|text|做这件事的好方法是有一个城市课程：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|371927|class+City{
++++private+int+id;
++++private+long+population;

++++//...+getters,+setters,+etc
}|code-block|syntax|javascript|371928|城市比较级：|371929|class+CityPopulationComparator+implements+Comparator<City>+{
++++@Override
++++public+int+compare(City+c1,+City+c2)+{
++++++++return+Long.compare(c1.getPopulation(),+c2.getPopulation());
++++}
}|371930|以及一个城市的数组列表：|371931|ArrayList<City>+cities;|371932|最后，使用以下方法对其进行排序：|371933|Collections.sort(cities,+new+CityPopulationComparator());|371934|但是--如果您需要以这种方式拥有您的城市和人口，您可以自己编写一个排序方法(例如，一个气泡排序)，并且每当您交换两个城市时，也可以交换相应的填充。|offset|length|style|BOLD|371935|entityMap^0|0|0|0|0|0|0|0|0|0|2|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|18|8|@$U|19|V|1A|W|X]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

The good way of doing this is having a city class:

<pre><code>class City{
 private int id;
 private long population;

 //... getters, setters, etc
}
</code></pre>

a city comparator class:

<pre><code>class CityPopulationComparator implements Comparator&lt;City&gt; {
 @Override
 public int compare(City c1, City c2) {
 return Long.compare(c1.getPopulation(), c2.getPopulation());
 }
}
</code></pre>

And an array list of cities:

<pre><code>ArrayList&lt;City&gt; cities;
</code></pre>

and finally sort it using:

<pre><code>Collections.sort(cities, new CityPopulationComparator());
</code></pre>

But if you need to have your cities and populations this way, you can write a sort method yourself (a bubble sort for example) and whenever you swap two cities, also swap corresponding pupulations.

blocks|key|1127518|text|如果您的城市标识是唯一的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1127519|int[]+city+=+{+3,++++1,+++5};
int[]+pop++=+{3333,+333,+33};
Map<Integer,+Integer>+arr+=+new+HashMap<>(city.length);
for+(int+i+=+0;+i+<+city.length;+i%2B%2B)+{
++++arr.put(city[i],+pop[i]);
}
Arrays.sort(city);
for+(int+i+=+0;+i+<+city.length;+i%2B%2B)+{
++++pop[i]+=+arr.get(city[i]);
}|code-block|syntax|javascript|1127520|同样使用SortedMap|offset|length|1127521|int[]+city+=+{+3,++++1,+++5};
int[]+pop++=+{3333,+333,+33};
SortedMap<Integer,+Integer>+arr+=+new+TreeMap<>();
for+(int+i+=+0;+i+<+city.length;+i%2B%2B)+{
++++arr.put(city[i],+pop[i]);
}
System.out.println(arr.keySet());
System.out.println(arr.values());|1127522|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/util/SortedMap.html^0|0|0|4|9|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

If your city id is unique:
<pre class="lang-java prettyprint-override"><code>int[] city = { 3, 1, 5};
int[] pop = {3333, 333, 33};
Map&lt;Integer, Integer&gt; arr = new HashMap&lt;&gt;(city.length);
for (int i = 0; i &lt; city.length; i++) {
 arr.put(city[i], pop[i]);
}
Arrays.sort(city);
for (int i = 0; i &lt; city.length; i++) {
 pop[i] = arr.get(city[i]);
}
</code></pre>
The same using <a href="https://docs.oracle.com/javase/7/docs/api/java/util/SortedMap.html" rel="nofollow noreferrer">SortedMap</a>
<pre class="lang-java prettyprint-override"><code>int[] city = { 3, 1, 5};
int[] pop = {3333, 333, 33};
SortedMap&lt;Integer, Integer&gt; arr = new TreeMap&lt;&gt;();
for (int i = 0; i &lt; city.length; i++) {
 arr.put(city[i], pop[i]);
}
System.out.println(arr.keySet());
System.out.println(arr.values());
</code></pre>

blocks|key|1127546|text|正确的解决方案是这。然而，如果你想要一个完全疯狂的黑客，你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1127547|public+final+class+ParallelIntArrays+extends+AbstractList<int[]>+{

++++private+final+int[]+array1;
++++private+final+int[]+array2;

++++public+ParallelIntArrays(int[]+array1,+int[]+array2)+{
++++++++if+(array1.length+!=+array2.length)
++++++++++++throw+new+IllegalArgumentException();
++++++++this.array1+=+array1;
++++++++this.array2+=+array2;
++++}

++++@Override
++++public+int[]+get(int+i)+{
++++++++return+new+int[]+{+array1[i],+array2[i]+};
++++}

++++@Override
++++public+int+size()+{
++++++++return+array1.length;
++++}

++++@Override
++++public+int[]+set(int+i,+int[]+a)+{
++++++++if+(a.length+!=+2)
++++++++++++throw+new+IllegalArgumentException();
++++++++int[]+b+=+get(i);
++++++++array1[i]+=+a[0];
++++++++array2[i]+=+a[1];
++++++++return+b;
++++}
}|code-block|syntax|javascript|1127548|然后你就可以：|1127549|int[]+city+=+{5,+++1,++2,++++4,+++3+++};
int[]+pop+=++{100,+30,+4000,+400,+5000};
new+ParallelIntArrays(city,+pop).sort(Comparator.comparingInt(arr+->+arr[0]));
System.out.println(Arrays.toString(city));
System.out.println(Arrays.toString(pop));|1127550|注意，正如上面所写的，ParallelIntArrays不能作为一个List正确地工作。例如，list.contains(list.get(0))会给false。如果您将其改为List<IntBuffer>或List<List<Integer>>，则会修复它。|style|CODE|1127551|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/36440636/3973077^0|8|1|0|0|0|0|0|B|H|Y|4|1B|Q|23|5|2G|F|2W|J|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|15|8|@$A|16|B|17|O|P]|$A|18|B|19|O|P]|$A|1A|B|1B|O|P]|$A|1C|B|1D|O|P]|$A|1E|B|1F|O|P]|$A|1G|B|1H|O|P]]|9|@]|C|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

The correct solution is <a href="https://stackoverflow.com/a/36440636/3973077">this</a>. However if you want a completely mad hack, you can do this:

<pre><code>public final class ParallelIntArrays extends AbstractList&lt;int[]&gt; {

 private final int[] array1;
 private final int[] array2;

 public ParallelIntArrays(int[] array1, int[] array2) {
 if (array1.length != array2.length)
 throw new IllegalArgumentException();
 this.array1 = array1;
 this.array2 = array2;
 }

 @Override
 public int[] get(int i) {
 return new int[] { array1[i], array2[i] };
 }

 @Override
 public int size() {
 return array1.length;
 }

 @Override
 public int[] set(int i, int[] a) {
 if (a.length != 2)
 throw new IllegalArgumentException();
 int[] b = get(i);
 array1[i] = a[0];
 array2[i] = a[1];
 return b;
 }
}
</code></pre>

Then you can do:

<pre><code>int[] city = {5, 1, 2, 4, 3 };
int[] pop = {100, 30, 4000, 400, 5000};
new ParallelIntArrays(city, pop).sort(Comparator.comparingInt(arr -&gt; arr[0]));
System.out.println(Arrays.toString(city));
System.out.println(Arrays.toString(pop));
</code></pre>

Note that as written above, <code>ParallelIntArrays</code> does not function correctly as a <code>List</code>. For example <code>list.contains(list.get(0))</code> would give <code>false</code>. If you made it a <code>List&lt;IntBuffer&gt;</code> or a <code>List&lt;List&lt;Integer&gt;&gt;</code> instead, it would be fixed.

blocks|key|1127555|text|public+static+void+main(String[]+args)+{
++++Integer[]+bidAmount+=+{3000,+54000,+2000,+1000,+5600};
++++String[]+bidderName+=+{"Jenny",+"Mike",+"Alvin",+"Berry",+"John"};
++++Map<Integer,+String>+stringTreeMap+=+new+TreeMap<>();
++++stringTreeMap.put(bidAmount[0],+bidderName[0]);
++++stringTreeMap.put(bidAmount[1],+bidderName[1]);
++++stringTreeMap.put(bidAmount[2],+bidderName[2]);
++++stringTreeMap.put(bidAmount[3],+bidderName[3]);
++++stringTreeMap.put(bidAmount[4],+bidderName[4]);

++++for+(Map.Entry<Integer,+String>+entry+:+stringTreeMap.entrySet())+{
++++++++Integer+key+=+entry.getKey();
++++++++String+value+=+entry.getValue();

++++++++System.out.println(key+%2B+"+=>+"+%2B+value);
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1127556|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-java prettyprint-override"><code>public static void main(String[] args) {
 Integer[] bidAmount = {3000, 54000, 2000, 1000, 5600};
 String[] bidderName = {&quot;Jenny&quot;, &quot;Mike&quot;, &quot;Alvin&quot;, &quot;Berry&quot;, &quot;John&quot;};
 Map&lt;Integer, String&gt; stringTreeMap = new TreeMap&lt;&gt;();
 stringTreeMap.put(bidAmount[0], bidderName[0]);
 stringTreeMap.put(bidAmount[1], bidderName[1]);
 stringTreeMap.put(bidAmount[2], bidderName[2]);
 stringTreeMap.put(bidAmount[3], bidderName[3]);
 stringTreeMap.put(bidAmount[4], bidderName[4]);

 for (Map.Entry&lt;Integer, String&gt; entry : stringTreeMap.entrySet()) {
 Integer key = entry.getKey();
 String value = entry.getValue();

 System.out.println(key + &quot; =&gt; &quot; + value);
 }
}
</code></pre>

I have two arrays, one stores the distance of the cities and the other stores the corresponding population. Everything works fine if the distance of the cities is in ascending order.
But let say if someone inputs the distance randomly. How can I sort the cities array and also make sure that the population of the respective city is in the same index as the index of its respective city population.
For example:
<ul>
<li>City 1 has population 333</li>
<li>City 3 has population 33333</li>
<li>City 5 has population 33</li>
</ul>
<pre class="lang-java prettyprint-override"><code>int[] city = {1, 3, 5};
int[] pop = {333, 33333, 33};
</code></pre>
Everything works fine because the city array is sorted already.
But when I input:
<pre class="lang-java prettyprint-override"><code>int[] city = {3, 1, 5};
int[] pop = {3333, 333, 33};
</code></pre>
Big problem!
I want sort the array city and make sure that the population array has all its elements at the same index as their respective city.

Sorting two parallel arrays

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。

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我有两个数组，一个存储城市的距离，另一个存储相应的人口。如果城市的距离是按上升顺序排列的，一切都很好。但是如果有人随机输入距离的话。如何对城市数组进行排序，并确保相应城市的人口与其各自城市人口的索引相同。例如：第一城市人口333人。城市3有33333人口。城市5有人口33int[] city = {1, 3, 5};i...

问排序两个并行阵列
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