Detect duplicate values in primitive Java array

问题

I want to detect duplicate values in a Java array. For example:

int[] array = { 3, 3, 3, 1, 5, 8, 11, 4, 5 };

How could I get the specific duplicated entry and how many times it occurs?

回答1:

I'll have a Map<Integer, Integer> where the first integer is the value of the number that occurs in the array and the second integer is the count (number of occurrence).

Run through the array.length in a loop
for each item in the array, do a map.containsKey(array[i]). If there exists a number in a map, increment that number (something like map.put(array[i], map.get(array[i]) + 1). Otherwise, create a new entry in a map (e.g map.put(array[i], 1).
Finally, iterate through the map and retrieve all keys where value is greater than 1.

回答2:

Seems like a job for data structure called multiset.

Multiset<Integer> mp = HashMultiset.create();
mp.addAll(Arrays.asList(new Integer[] { 3, 3, 3, 1, 5, 8, 11, 4, 5 }));

Standard JDK 6 is primitive and do not contain multiset. If you do not want to rewrite it, you can use preexisting library like Google Guava-libraries or Apache Commons.

For example with Guava-libraries you can

    for (Integer i : mp.elementSet())
        System.out.println(i + " is contained " + mp.count(i) + " times.");

And this would output:

1 is contained 1 times.
3 is contained 3 times.
4 is contained 1 times.
5 is contained 2 times.
8 is contained 1 times.
11 is contained 1 times.

回答3:

The answer depends on the number range in your source array. If the range is small enough you can allocate an array, loop through your source and increment at the index of your source number:

int[] counts = new int[max_value + 1];

for (int n: array) {
    counts[n]++;
}

If your source array contains an unknown or too large range, you can create a Map and count in that:

Map<Integer,Integer> counts = new HashMap<Integer,Integer>();

for (Integer n: array) {
    if (counts.containsKey(n)) {
        counts.put(n, counts.get(n) + 1);
    } else {
        counts.put(n, 1);
    }
}

NB. typed the above without the help of a JVM, getting rid of typoes is left as an exercise for the reader :-)

回答4:

public class Duplicate {

    public static void main(String[] arg) {
        int[] array = {1, 3, 5, 6, 2, 3, 6, 4, 3, 2, 1, 6, 3};

        displayDuplicate(array);

    }

    static void displayDuplicate(int[] ar) {
        boolean[] done = new boolean[ar.length];
        for(int i = 0; i < ar.length; i++) {
            if(done[i])
                continue;
            int nb = 0;
            for(int j = i; j < ar.length; j++) {
                if(done[j])
                    continue;
                if(ar[j] == ar[i]) {
                    done[j] = true;
                    nb++;
                }
            }
            System.out.println(ar[i] + " occurs " + nb + " times");
        }
    }
}

回答5:

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;

public class DuplicatedValuesInArray 
{

    public static void main(String args[]) {  
        int[] array = { 3, 3, 3, 1, 5, 8, 11, 4, 5 };
        Map<Integer, Integer> map= new HashMap<Integer, Integer>();

      for(int i=0;i<array.length;i++) {   
          if(map.containsKey(array[i]))

          map.put(array[i],map.get(array[i]) + 1);
      else
          map.put(array[i], 1);
      }

      for (Integer i : map.keySet()) {
          System.out.println(i + " is contained " + map.get(i) + " times.");
      }
   }
}

回答6:

You can use Collectors.frecuency() and Collectors.groupingBy.

This is how i do this, hope this can help you.

    Map<Object,Long> finalValues = new HashMap<Object, Long>();

    finalValues = Arrays.asList(new Integer[] {3, 3, 3, 1, 5, 8, 11, 4, 5 })
            .stream()
            .collect(Collectors.groupingBy(e -> e, Collectors.counting()));

    //Check the output
    finalValues.entrySet().forEach(entry -> {
        System.out.println("number: " + entry.getKey() + "| Times: "+ entry.getValue());
    });

The output is:

number: 1| Times: 1
number: 3| Times: 3
number: 4| Times: 1
number: 5| Times: 2
number: 8| Times: 1
number: 11| Times: 1

You can even use frecuency to delete all the numbers that doesn't repeat:

Map finalValues = new HashMap();

    List<Integer> numbers = Arrays.asList(new Integer[]{1,2,1,3,4,4});     

    finalValues = numbers
            .stream()
            .filter(x-> Collections.frequency(numbers, x) > 1)
            .collect(Collectors.groupingBy(e -> e, Collectors.counting()));

    //Check the output
    finalValues.entrySet().forEach(entry -> {
        System.out.println("number: " + entry.getKey() + "| Times: "+ entry.getValue());
    });

The output is :

number: 1| Times: 2
number: 4| Times: 2

回答7:

assign a counter for the first step then you can relate them to an another array assigning each number to an index then if your number is duplicated increment your counter...

回答8:

Sort the array, then either scan it or Arrays.binarySearch + scan in either direction. Due to much fewer allocations and no wrapping, this can be faster, especially on larger arrays.

回答9:

 Java 8, the solution:
1. Create Map when the Key is the Value of Array and Value is counter.
2. Check if Map contains the Key increase counter or add a new set.
private static void calculateDublicateValues(int[] array) {
      //key is value of array, value is counter
      Map<Integer, Integer> map = new HashMap<Integer, Integer>();

      for (Integer element : array) {
        if (map.containsKey(element)) {
          map.put(element, map.get(element) + 1); // increase counter if contains
        } else
          map.put(element, 1);
      }

      map.forEach((k, v) -> {
        if (v > 1)
          System.out.println("The element " + k + " duplicated " + v + " times");
      });

    }

来源：https://stackoverflow.com/questions/5364154/detect-duplicate-values-in-primitive-java-array

标签

java

arrays

duplicates

primitive