确定每个整数出现在输入（数组）中的次数

Question

it keeps repeating regardless of whether it's been already calculated or not. You can see the example output. it already calculated the occurrence of 1, but when it sees 1 again, it will calculate it again!

public class SortingInLinearTime {
    public static int[][] howMany(int[] n){

        // Makes a double array list where the second value is occurrence of the first value.
        int[][] a = new int[n.length][2]; 
        int counter = 0;

        for(int i = 0; i != n.length; i++){
            for(int j = 0; j != n.length; j++) {

                if(n[i] == n[j]){
                    counter++;
                }
            }
            a[i][0] = n[i];
            a[i][1] = counter;
            counter = 0;
        }

        // printer helper function
        for(int i = 0; i != n.length; i++){ 
            System.out.print(a[i][0] + " occurs ");
            System.out.println(a[i][1] + " times");
        }
        return a;
    }

    public static void main(String[] args) {
        int[] testArray = {1, 2, 3, 1, 2, 3, 4};
        System.out.print(howMany(testArray));
    }
}

output: 1 occurs 2 times 2 occurs 2 times 3 occurs 2 times 1 occurs 2 times 2 occurs 2 times 3 occurs 2 times 4 occurs 1 times [[I@15db9742

Answer 1

In the first loop with i, you are recounting the same values again and again. 1 appears when i = 0 and when i = 3 as well. you once counted for 1 when i == 0 and recounted again at i == 3 in array n. However, I believe the best solution for your problem could be achieved by changing your data structure from int[][] to an hashmap.

Answer 2

Convert your array to list using Arrays.asList() and then use the collections api to get the count.

Collections.frequency(Collection c, Object o)

Updated with the implementation

import java.util.AbstractList;
import java.util.Collections;
import java.util.List;

public class SortingInLinearTime {
    public static int[][] howMany( int[] n){

        // Makes a double array list where the second value is occurrence of the first value.
        int[][] a = new int[n.length][2]; 


        for(int i = 0; i < n.length; i++){
            int count = Collections.frequency(asList(n), n[i]);
            a[i][0] = n[i];
            a[i][1] = count;
         }

        // printer helper function
        for(int i = 0; i < n.length; i++){ 
            System.out.print(a[i][0] + " occurs ");
            System.out.println(a[i][1] + " times");
        }
        return a;
    }


    public static List<Integer> asList(final int[] is)
    {
        return new AbstractList<Integer>() {
                public Integer get(int i) { return is[i]; }
                public int size() { return is.length; }
        };
    }


    public static void main(String[] args) {
        int[] testArray = {1, 2, 3, 1, 2, 3, 4};
        System.out.print(howMany(testArray));
    }
}

Answer 3

You're making this way harder than it needs to be both by looping over the array twice and by using an array to store your result. Try this:

public class SortingInLinearTime {
    public static Hashtable<Integer, Integer> howMany(int[] n){
        Hashtable<Integer, Integer> toRet = new Hashtable<Integer, Integer>();
        for (int i = 0; i < n.length; i++) {
            if (!toRet .containsKey(n[i])) {
                toRet.put(n[i], 1);
            } else {
                toRet.put(n[i], toRet.get(n[i]) + 1);
            }
        }
        return toRet;
    }

    public static void main(String[] args) {
        int[] testArray = {1, 2, 3, 1, 2, 3, 4};
        Hashtable<Integer, Integer> counts = howMany(testArray);
        Set<Integer> keys = counts.keySet();
        for(Integer key : keys){
            System.out.println(key + " occurs " + counts.get(key) + " times.");
        }
    }
}

This has several advantages. It will not break if you pass an array with large numbers, like {1, 11, 203, 203} , which your current implementation cannot handle. It does not use extra space by declaring an array with many elements that you do not need. Most important, it works.