CODEWITHSUNDEEP
javaalgorithmbubble-sort
user20758216
user20758216

Reputation:

Implementing a swap method in a bubble sort

I have a program that I have written that allows the user to enter some votes into a system. In the end, the number entered correlates to the person of that number. (0 = Ahmed, 1 = boo, etc)

(current)At the end of the voting period the program will sort the votes in descending order and output them (once the user has entered -1).

(future) the name of the candidate should be shown in this sorted list, e.g. ahmed - 5, celine - 4, didi -3, boo - 0, elaine - 0.

I have the code writing the sorted vote numbers, but I am struggling to understand how it works to implement the swap with the name array to produce the final result as listed before in future. Can anyone help with this? Below is my code:

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        int[] NumArray = new int[5];
        final int EXIT_CODE = -1;
        displayInitialText();
        int userInput = input.nextInt();
        while (userInput != EXIT_CODE) {
            if (userInput >= 0 && userInput <= 4) {
                voteNumArray[userInput] = voteNumArray[userInput] + 1;
            } else {
                System.out.println("\n-----------------------");
                System.out.println("Invalid Vote - Please enter a number between 0-4");
                System.out.println("-----------------------\n");
            }
    
            userInput = input.nextInt();
        }
        sort(voteNumArray);
        displaySortedItems(voteNumArray);
    }

Upvotes: 2

Views: 324

Answers (2)

user19443549
user19443549

Reputation:

Bubble sort compares adjacent values and swaps if not in desired order so after every pass you will get largest/minimum(depending on implementation) value at the end of array.

let assume your array is having below values and you are sorting in ascending order -

2,5,1,7,4

In First pass, i = 0 -

  1. It will compare 2,5 and 5 is greater than 2, so no swap.

  2. then it will compare 5,1 and 1 is smaller so it will swap and your array will become as below -

    2,1,5,7,4

  3. compare 5,7 and no swap as 7 is greater.

  4. compare 7,4 and swap as 4 is smaller and your array will become -

    2,5,1,4,7

After second pass, i = 1 -

2,1,4,5,7

After third pass, i = 2 -

1,2,4,5,7

Your array is sorted now, but still it will complete rest 2 passes but wont do any swap.

In every iteration you gets largest(considering ascending sorting) number in the end and thats the reason you are doing j < array.length - i -1, because you just need to sort remaining array.

Upvotes: 0

trincot
trincot

Reputation: 350137

You would need a second swap function:

    public static void main(String[] args) {
        // ... ...
        // ... ...

        sort(voteNumArray, voteNameArray); // Pass the second array too
        displaySortedItems(voteNumArray);
     }
     public static void sort(int[] array, String[] names){
        for (int i = 0; i < (array.length-1);i++){
            for(int j = 0; j < (array.length - i -1);j++){
                if(array[j] < array[j+1]){
                    swapInts(array,j);
                    swapStrings(names,j); // Swap in second array too
                }
            }
        }
    }
    public static void swapInts(int[] array, int index){
        int temp = array[index];
        array[index] = array[index + 1];
        array[index + 1] = temp;
    }
    public static void swapStrings(String[] array, int index){ // Extra swap function
        String temp = array[index];
        array[index] = array[index + 1];
        array[index + 1] = temp;
    }

Upvotes: 1

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