blocks|key|430183|text|使用Collections.frequency选项：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|430184|+List<String>+list+=+Arrays.asList("1",+"1","1","1","1","1","5","5","12","12","12","12","12","12","12","12","12","12","8");
++++++int+max+=+0;
++++++int+curr+=+0;
++++++String+currKey+=++null;
++++++Set<String>+unique+=+new+HashSet<String>(list);

++++++++++for+(String+key+:+unique)+{
++++++++++++++++curr+=+Collections.frequency(list,+key);

+++++++++++++++if(max+<+curr){
+++++++++++++++++max+=+curr;
+++++++++++++++++currKey+=+key;
++++++++++++++++}
++++++++++++}

++++++++++System.out.println("The+number+"++%2B+currKey+%2B+"+happens+"+%2B+max+%2B+"+times");|code-block|syntax|javascript|430185|输出：|430186|The+number+12+happens+10+times|430187|entityMap^0|2|L|0|0|0|0|U|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

Use <code>Collections.frequency</code> option:

<pre><code> List&lt;String&gt; list = Arrays.asList("1", "1","1","1","1","1","5","5","12","12","12","12","12","12","12","12","12","12","8");
 int max = 0;
 int curr = 0;
 String currKey = null;
 Set&lt;String&gt; unique = new HashSet&lt;String&gt;(list);

 for (String key : unique) {
 curr = Collections.frequency(list, key);

 if(max &lt; curr){
 max = curr;
 currKey = key;
 }
 }

 System.out.println("The number " + currKey + " happens " + max + " times");
</code></pre>

Output:

<code>The number 12 happens 10 times</code>

blocks|key|211902|text|使用Java+8的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|211903|+++++++int+result+=+Arrays.stream(array)
+++++++++++.boxed()
+++++++++++.collect(Collectors.groupingBy(i->i,Collectors.counting()))
+++++++++++.values()
+++++++++++.stream()
+++++++++++.max(Comparator.comparingLong(i->i))
+++++++++++.orElseThrow(RuntimeException::new));|code-block|syntax|javascript|211904|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The solution with Java 8

<pre><code> int result = Arrays.stream(array)
 .boxed()
 .collect(Collectors.groupingBy(i-&gt;i,Collectors.counting()))
 .values()
 .stream()
 .max(Comparator.comparingLong(i-&gt;i))
 .orElseThrow(RuntimeException::new));
</code></pre>

blocks|key|211726|text|我将建议另一种方法。我不知道这样做会不会更快。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|211727|对数组进行快速排序。使用内置的Arrays.sort()方法。|211728|现在比较相邻的元素。考虑这个例子：|211729|1+1+1+4+4+4+9+9+9+10+10+29+29+29|211730|当相邻元素不相等时，您可以停止计算该元素。|211731|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|O|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

I will suggest another method. I don't know if this would work faster or not.

Quick sort the array. Use the built in Arrays.sort() method.

Now compare the adjacent elements. 
Consider this example:

1 1 1 1 4 4 4 4 4 4 4 4 4 4 4 4 9 9 9 10 10 10 29 29 29 29 29 29

When the adjacent elements are not equal, you can stop counting that element.

blocks|key|211866|text|解决方案1:使用HashMap|type|unstyled|depth|inlineStyleRanges|entityRanges|data|211867|class+test1+{
++++public+static+void+main(String[]+args)+{

++++int[]+a+=+{1,1,2,1,5,6,6,6,8,5,9,7,1};
++++//+max+occurences+of+an+array
++++Map<Integer,Integer>+map+=+new+HashMap<>();
++++++int+max+=+0+;+int+chh+=+0+;
++++++for(int+i+=+0+;+i+<+a.length;i%2B%2B)+{
++++++++++int+ch+=+a[i];
++++++++++map.put(ch,+map.getOrDefault(ch,+0)+%2B1);
++++++}//for
++++++Set<Entry<Integer,Integer>>+entrySet+=map.entrySet();

++++++for(Entry<Integer,Integer>+entry+:+entrySet)+{
++++++++++if(entry.getValue()+>+max)+{max+=+entry.getValue();chh+=+entry.getKey();}

++++++}//for
++++System.out.println("max+element+=>+"+%2B+chh);
++++System.out.println("frequency+=>+"+%2B+max);
++++}//amin
}
/*output+=>
max+element+=>+1
frequency+=>+4

*/|code-block|syntax|javascript|211868|解决方案2:使用计数数组|211869|public+class+test2+{
++++public+static+void+main(String[]+args)+{
+++++++++int[]+a+=+{1,1,2,1,5,6,6,6,6,6,8,5,9,7,1};
++++int+max+=+0+;+int+chh+=+0;
++++int+count[]+=+new+int[a.length];
++++for(int+i+=+0+;+i+<a.length+;+i%2B%2B)+{
++++++++int+ch+=+a[i];
++++++++count[ch]+%2B=1+;
++++}//for

++++for(int+i+=+0+;+i+<a.length+;i%2B%2B)++{
++++++++int+ch+=+a[i];
++++++++if(count[ch]+>+max)+{max+=+count[ch]+;+chh+=+ch+;}
++++}//for
+++++++++System.out.println(chh);+
++++}//main
}|211870|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Solution 1: Using HashMap

<pre><code>class test1 {
 public static void main(String[] args) {

 int[] a = {1,1,2,1,5,6,6,6,8,5,9,7,1};
 // max occurences of an array
 Map&lt;Integer,Integer&gt; map = new HashMap&lt;&gt;();
 int max = 0 ; int chh = 0 ;
 for(int i = 0 ; i &lt; a.length;i++) {
 int ch = a[i];
 map.put(ch, map.getOrDefault(ch, 0) +1);
 }//for
 Set&lt;Entry&lt;Integer,Integer&gt;&gt; entrySet =map.entrySet();

 for(Entry&lt;Integer,Integer&gt; entry : entrySet) {
 if(entry.getValue() &gt; max) {max = entry.getValue();chh = entry.getKey();}

 }//for
 System.out.println("max element =&gt; " + chh);
 System.out.println("frequency =&gt; " + max);
 }//amin
}
/*output =&gt;
max element =&gt; 1
frequency =&gt; 4

*/
</code></pre>

Solution 2 : Using count array

<pre><code>public class test2 {
 public static void main(String[] args) {
 int[] a = {1,1,2,1,5,6,6,6,6,6,8,5,9,7,1};
 int max = 0 ; int chh = 0;
 int count[] = new int[a.length];
 for(int i = 0 ; i &lt;a.length ; i++) {
 int ch = a[i];
 count[ch] +=1 ;
 }//for

 for(int i = 0 ; i &lt;a.length ;i++) {
 int ch = a[i];
 if(count[ch] &gt; max) {max = count[ch] ; chh = ch ;}
 }//for
 System.out.println(chh); 
 }//main
}
</code></pre>

blocks|key|430376|text|这是一个java解决方案--|type|unstyled|depth|inlineStyleRanges|entityRanges|data|430377|++++++++List<Integer>+list+=+Arrays.asList(1,+2,+2,+3,+2,+1,+3);
++++++++Set<Integer>+set+=+new+HashSet(list);
++++++++int+max+=+0;
++++++++int+maxtemp;
++++++++int+currentNum+=+0;
++++++++for+(Integer+k+:+set)+{
++++++++++++maxtemp+=+Math.max(Collections.frequency(list,+k),+max);
++++++++++++currentNum+=+maxtemp+!=+max+?+k+:+currentNum;
++++++++++++max+=+maxtemp;
++++++++}
++++++++System.out.println("Number+::+"+%2B+currentNum+%2B+"+Occurs+::+"+%2B+max+%2B+"+times");|code-block|syntax|javascript|430378|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's a java solution --
<pre class="lang-java prettyprint-override"><code> List&lt;Integer&gt; list = Arrays.asList(1, 2, 2, 3, 2, 1, 3);
 Set&lt;Integer&gt; set = new HashSet(list);
 int max = 0;
 int maxtemp;
 int currentNum = 0;
 for (Integer k : set) {
 maxtemp = Math.max(Collections.frequency(list, k), max);
 currentNum = maxtemp != max ? k : currentNum;
 max = maxtemp;
 }
 System.out.println(&quot;Number :: &quot; + currentNum + &quot; Occurs :: &quot; + max + &quot; times&quot;);
</code></pre>

blocks|key|211945|text|int[]+array+=+new+int[]+{+1,+2,+4,+1,+3,+4,+2,+2,+1,+5,+2,+3,+5+};

Long+max+=+Arrays.stream(array).boxed().collect(Collectors.groupingBy(i+->+i,+Collectors.counting())).values()
++++++++++++++++.stream().max(Comparator.comparing(Function.identity())).orElse(0L);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|211946|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><code>int[] array = new int[] { 1, 2, 4, 1, 3, 4, 2, 2, 1, 5, 2, 3, 5 };

Long max = Arrays.stream(array).boxed().collect(Collectors.groupingBy(i -&gt; i, Collectors.counting())).values()
 .stream().max(Comparator.comparing(Function.identity())).orElse(0L);
 
</code></pre>

blocks|key|211752|text|public+static+void+main(String[]+args)+{

+++int+n;

+++int[]+arr;

++++Scanner+in+=+new+Scanner(System.in);
++++System.out.println("Enter+Length+of+Array");
++++n+=+in.nextInt();
++++arr+=+new+int[n];
++++System.out.println("Enter+Elements+in+array");+

++++for+(int+i+=+0;+i+<+n;+i%2B%2B)+{
++++++++arr[i]+=+in.nextInt();
++++}

++++int+greatest+=+arr[0];

++++for+(int+i+=+0;+i+<+arr.length;+i%2B%2B)+{
++++++++if+(arr[i]+>+greatest)+{
++++++++++++greatest+=+arr[i];
++++++++}

++++}+

++++System.out.println("Greatest+Number+"+%2B+greatest);

++++int+count+=+0;

++++for+(int+i+=+0;+i+<+arr.length;+i%2B%2B)+{
++++++++if+(greatest+==+arr[i])+{
++++++++++++count%2B%2B;
++++++++}
++++}

++++System.out.println("Number+of+Occurance+of+"+%2B+greatest+%2B+":"+%2B+count+%2B+"+times");

++++in.close();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|211753|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><code>public static void main(String[] args) {

 int n;

 int[] arr;

 Scanner in = new Scanner(System.in);
 System.out.println("Enter Length of Array");
 n = in.nextInt();
 arr = new int[n];
 System.out.println("Enter Elements in array"); 

 for (int i = 0; i &lt; n; i++) {
 arr[i] = in.nextInt();
 }

 int greatest = arr[0];

 for (int i = 0; i &lt; arr.length; i++) {
 if (arr[i] &gt; greatest) {
 greatest = arr[i];
 }

 } 

 System.out.println("Greatest Number " + greatest);

 int count = 0;

 for (int i = 0; i &lt; arr.length; i++) {
 if (greatest == arr[i]) {
 count++;
 }
 }

 System.out.println("Number of Occurance of " + greatest + ":" + count + " times");

 in.close();
}
</code></pre>

blocks|key|211817|text|为了继续你写的伪代码，试试下面写的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|211818|public+static+void+fetchFrequency(int[]+arry)+{
++++++++Map<Integer,+Integer>+newMap+=+new+TreeMap<Integer,+Integer>(Collections.reverseOrder());
++++++++int+num+=+0;
++++++++int+count+=+0;
++++++++for+(int+i+=+0;+i+<+arry.length;+i%2B%2B)+{
++++++++++++if+(newMap.containsKey(arry[i]))+{
++++++++++++++++count+=+newMap.get(arry[i]);
++++++++++++++++newMap.put(arry[i],+%2B%2Bcount);
++++++++++++}+else+{
++++++++++++++++newMap.put(arry[i],+1);
++++++++++++}
++++++++}
++++++++Set<Entry<Integer,+Integer>>+set+=+newMap.entrySet();
++++++++List<Entry<Integer,+Integer>>+list+=+new+ArrayList<Entry<Integer,+Integer>>(set);
++++++++Collections.sort(list,+new+Comparator<Map.Entry<Integer,+Integer>>()+{

++++++++++++@Override
++++++++++++public+int+compare(Entry<Integer,+Integer>+o1,+Entry<Integer,+Integer>+o2)+{
++++++++++++++++return+(o2.getValue()).compareTo(o1.getValue());
++++++++++++}
++++++++});

++++++++for+(Map.Entry<Integer,+Integer>+entry+:+list)+{
++++++++++++System.out.println(entry.getKey()+%2B+"+====+"+%2B+entry.getValue());
++++++++++++break;
++++++++}
++++++++//return+num;
++++}|code-block|syntax|javascript|211819|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In continuation to the pseudo-code what you've written try the below written code:-

<pre><code>public static void fetchFrequency(int[] arry) {
 Map&lt;Integer, Integer&gt; newMap = new TreeMap&lt;Integer, Integer&gt;(Collections.reverseOrder());
 int num = 0;
 int count = 0;
 for (int i = 0; i &lt; arry.length; i++) {
 if (newMap.containsKey(arry[i])) {
 count = newMap.get(arry[i]);
 newMap.put(arry[i], ++count);
 } else {
 newMap.put(arry[i], 1);
 }
 }
 Set&lt;Entry&lt;Integer, Integer&gt;&gt; set = newMap.entrySet();
 List&lt;Entry&lt;Integer, Integer&gt;&gt; list = new ArrayList&lt;Entry&lt;Integer, Integer&gt;&gt;(set);
 Collections.sort(list, new Comparator&lt;Map.Entry&lt;Integer, Integer&gt;&gt;() {

 @Override
 public int compare(Entry&lt;Integer, Integer&gt; o1, Entry&lt;Integer, Integer&gt; o2) {
 return (o2.getValue()).compareTo(o1.getValue());
 }
 });

 for (Map.Entry&lt;Integer, Integer&gt; entry : list) {
 System.out.println(entry.getKey() + " ==== " + entry.getValue());
 break;
 }
 //return num;
 }
</code></pre>

blocks|key|211892|text|这就是我在java中实现的方式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|211893|import+java.io.*;
class+Prog8
{
++++public+static+void+main(String[]+args)+throws+IOException+
++++{
++++++++BufferedReader+br=new+BufferedReader(new+InputStreamReader(System.in));
++++++++System.out.println("Input+Array+Size:");
++++++++int+size=Integer.parseInt(br.readLine());
++++++++int[]+arr=+new+int[size];
++++++++System.out.println("Input+Elements+in+Array:");
++++++++for(int+i=0;i<size;i%2B%2B)
++++++++++++arr[i]=Integer.parseInt(br.readLine());
++++++++int+max+=+0,pos=0,count+=+0;
++++++++for+(int+i+=+0;+i+<+arr.length;+i%2B%2B)
++++++++{
++++++++++++count=0;
++++++++++++for+(int+j+=+0;+j+<+arr.length;+j%2B%2B)+
++++++++++++{
++++++++++++++++if+(arr[i]==arr[j])
++++++++++++++++++++count%2B%2B;
++++++++++++}
++++++++++++if+(count+>=max)
++++++++++++{
++++++++++++++++max+=+count;
++++++++++++++++pos=i;
++++++++++++}
++++++++}

++++++++if(max==1)
++++++++++++System.out.println("No+Duplicate+Element.");
++++++++else
++++++++++++System.out.println("Element:"%2Barr[pos]%2B"+Occourance:"%2Bmax);
++++}
}|code-block|syntax|javascript|211894|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is how i have implemented in java..

<pre><code>import java.io.*;
class Prog8
{
 public static void main(String[] args) throws IOException 
 {
 BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
 System.out.println("Input Array Size:");
 int size=Integer.parseInt(br.readLine());
 int[] arr= new int[size];
 System.out.println("Input Elements in Array:");
 for(int i=0;i&lt;size;i++)
 arr[i]=Integer.parseInt(br.readLine());
 int max = 0,pos=0,count = 0;
 for (int i = 0; i &lt; arr.length; i++)
 {
 count=0;
 for (int j = 0; j &lt; arr.length; j++) 
 {
 if (arr[i]==arr[j])
 count++;
 }
 if (count &gt;=max)
 {
 max = count;
 pos=i;
 }
 }

 if(max==1)
 System.out.println("No Duplicate Element.");
 else
 System.out.println("Element:"+arr[pos]+" Occourance:"+max);
 }
}
</code></pre>

blocks|key|211921|text|下面是Ruby+SOlution：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|211922|def+maxOccurence(arr)
++m_hash+=+arr.group_by(&:itself).transform_values(&:count)
++elem+=+0,+elem_count+=+0
++m_hash.each+do+%7Ck,+v%7C
++++if+v+>+elem_count
++++++++elem+=+k
++++++++elem_count+=+v
++++end
++end
++"#{elem}+occured+#{elem_count}+times"
end

p+maxOccurence(["1",+"1","1","1","1","1","5","5","12","12","12","12","12","12","12","12","12","12","8"])|code-block|syntax|javascript|211923|输出：|211924|"12+occured+10+times"|211925|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here is Ruby SOlution:
<pre><code>def maxOccurence(arr)
 m_hash = arr.group_by(&amp;:itself).transform_values(&amp;:count)
 elem = 0, elem_count = 0
 m_hash.each do |k, v|
 if v &gt; elem_count
 elem = k
 elem_count = v
 end
 end
 &quot;#{elem} occured #{elem_count} times&quot;
end

p maxOccurence([&quot;1&quot;, &quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;5&quot;,&quot;5&quot;,&quot;12&quot;,&quot;12&quot;,&quot;12&quot;,&quot;12&quot;,&quot;12&quot;,&quot;12&quot;,&quot;12&quot;,&quot;12&quot;,&quot;12&quot;,&quot;12&quot;,&quot;8&quot;])
</code></pre>
output:
<pre><code>&quot;12 occured 10 times&quot;
</code></pre>

I have to find the element with highest occurrences in a double array.
I did it like this:

<pre><code>int max = 0;
for (int i = 0; i &lt; array.length; i++) {
 int count = 0;
 for (int j = 0; j &lt; array.length; j++) {
 if (array[i]==array[j])
 count++;
 }
 if (count &gt;= max)
 max = count;
 }
</code></pre>

The program works, but it is too slow! I have to find a better solution, can anyone help me?

Find the element with highest occurrences in an array [java]

Java

我必须找到双精度数组中出现次数最多的元素。我是这样做的：int max = 0;for (int i = 0; i < array.length; i++) { int count = 0; for (int j = 0; j < array.length; j++) { if (array[i]==array[j]) count++

问查找数组中出现次数最多的元素[java]
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回答 10

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回答 10

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问查找数组中出现次数最多的元素[java]
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