I have an array of objects in Java, and I am trying to pull one element to the top and shift the rest down by one.
Assume I have an array of size 10, and I am trying to pull the fifth element. The fifth element goes into position 0 and all elements from 0 to 5 will be shifted down by one.
This algorithm does not properly shift the elements:
Object temp = pool[position];
for (int i = 0; i < position; i++) {
array[i+1] = array[i];
}
array[0] = temp;
How do I do it correctly?
Logically it does not work and you should reverse your loop:
for (int i = position-1; i >= 0; i--) {
array[i+1] = array[i];
}
Alternatively you can use
System.arraycopy(array, 0, array, 1, position);
Assuming your array is {10,20,30,40,50,60,70,80,90,100}
What your loop does is:
Iteration 1: array[1] = array[0]; {10,10,30,40,50,60,70,80,90,100}
Iteration 2: array[2] = array[1]; {10,10,10,40,50,60,70,80,90,100}
What you should be doing is
Object temp = pool[position];
for (int i = (position - 1); i >= 0; i--) {
array[i+1] = array[i];
}
array[0] = temp;
You can just use Collections.rotate(List<?> list, int distance)
Use Arrays.asList(array) to convert to List
more info at: https://docs.oracle.com/javase/7/docs/api/java/util/Collections.html#rotate(java.util.List,%20int)
Instead of shifting by one position you can make this function more general using module like this.
int[] original = { 1, 2, 3, 4, 5, 6 };
int[] reordered = new int[original.length];
int shift = 1;
for(int i=0; i<original.length;i++)
reordered[i] = original[(shift+i)%original.length];
Just for completeness: Stream solution since Java 8.
final String[] shiftedArray = Arrays.stream(array)
.skip(1)
.toArray(String[]::new);
I think I sticked with the System.arraycopy() in your situtation. But the best long-term solution might be to convert everything to Immutable Collections (Guava, Vavr), as long as those collections are short-lived.
Manipulating arrays in this way is error prone, as you've discovered. A better option may be to use a LinkedList in your situation. With a linked list, and all Java collections, array management is handled internally so you don't have to worry about moving elements around. With a LinkedList you just call remove and then addLast and the you're done.
Try this:
Object temp = pool[position];
for (int i = position-1; i >= 0; i--) {
array[i+1] = array[i];
}
array[0] = temp;
Look here to see it working: http://www.ideone.com/5JfAg
Using array Copy
Generic solution for k times shift k=1 or k=3 etc
public void rotate(int[] nums, int k) {
// Step 1
// k > array length then we dont need to shift k times because when we shift
// array length times then the array will go back to intial position.
// so we can just do only k%array length times.
// change k = k% array.length;
if (k > nums.length) {
k = k % nums.length;
}
// Step 2;
// initialize temporary array with same length of input array.
// copy items from input array starting from array length -k as source till
// array end and place in new array starting from index 0;
int[] tempArray = new int[nums.length];
System.arraycopy(nums, nums.length - k, tempArray, 0, k);
// step3:
// loop and copy all the remaining elements till array length -k index and copy
// in result array starting from position k
for (int i = 0; i < nums.length - k; i++) {
tempArray[k + i] = nums[i];
}
// step 4 copy temp array to input array since our goal is to change input
// array.
System.arraycopy(tempArray, 0, nums, 0, tempArray.length);
}
code
public void rotate(int[] nums, int k) {
if (k > nums.length) {
k = k % nums.length;
}
int[] tempArray = new int[nums.length];
System.arraycopy(nums, nums.length - k, tempArray, 0, k);
for (int i = 0; i < nums.length - k; i++) {
tempArray[k + i] = nums[i];
}
System.arraycopy(tempArray, 0, nums, 0, tempArray.length);
}
In the first iteration of your loop, you overwrite the value in array[1]. You should go through the indicies in the reverse order.
static void pushZerosToEnd(int arr[])
{ int n = arr.length;
int count = 0; // Count of non-zero elements
// Traverse the array. If element encountered is non-zero, then
// replace the element at index 'count' with this element
for (int i = 0; i < n; i++){
if (arr[i] != 0)`enter code here`
// arr[count++] = arr[i]; // here count is incremented
swapNumbers(arr,count++,i);
}
for (int j = 0; j < n; j++){
System.out.print(arr[j]+",");
}
}
public static void swapNumbers(int [] arr, int pos1, int pos2){
int temp = arr[pos2];
arr[pos2] = arr[pos1];
arr[pos1] = temp;
}
Another variation if you have the array data as a Java-List
listOfStuff.add(
0,
listOfStuff.remove(listOfStuff.size() - 1) );
Just sharing another option I ran across for this, but I think the answer from #Murat Mustafin is the way to go with a list
public class Test1 {
public static void main(String[] args) {
int[] x = { 1, 2, 3, 4, 5, 6 };
Test1 test = new Test1();
x = test.shiftArray(x, 2);
for (int i = 0; i < x.length; i++) {
System.out.print(x[i] + " ");
}
}
public int[] pushFirstElementToLast(int[] x, int position) {
int temp = x[0];
for (int i = 0; i < x.length - 1; i++) {
x[i] = x[i + 1];
}
x[x.length - 1] = temp;
return x;
}
public int[] shiftArray(int[] x, int position) {
for (int i = position - 1; i >= 0; i--) {
x = pushFirstElementToLast(x, position);
}
return x;
}
}
A left rotation operation on an array of size n shifts each of the array's elements unit to the left, check this out!!!!!!
public class Solution {
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
String[] nd = scanner.nextLine().split(" ");
int n = Integer.parseInt(nd[0]); //no. of elements in the array
int d = Integer.parseInt(nd[1]); //number of left rotations
int[] a = new int[n];
for(int i=0;i<n;i++){
a[i]=scanner.nextInt();
}
Solution s= new Solution();
//number of left rotations
for(int j=0;j<d;j++){
s.rotate(a,n);
}
//print the shifted array
for(int i:a){System.out.print(i+" ");}
}
//shift each elements to the left by one
public static void rotate(int a[],int n){
int temp=a[0];
for(int i=0;i<n;i++){
if(i<n-1){a[i]=a[i+1];}
else{a[i]=temp;}
}}
}
You can use the Below codes for shifting not rotating:
int []arr = {1,2,3,4,5,6,7,8,9,10,11,12};
int n = arr.length;
int d = 3;
Programm for shifting array of size n by d elements towards left:
Input : {1,2,3,4,5,6,7,8,9,10,11,12}
Output: {4,5,6,7,8,9,10,11,12,10,11,12}
public void shiftLeft(int []arr,int d,int n) {
for(int i=0;i<n-d;i++) {
arr[i] = arr[i+d];
}
}
Programm for shifting array of size n by d elements towards right:
Input : {1,2,3,4,5,6,7,8,9,10,11,12}
Output: {1,2,3,1,2,3,4,5,6,7,8,9}
public void shiftRight(int []arr,int d,int n) {
for(int i=n-1;i>=d;i--) {
arr[i] = arr[i-d];
}
}
import java.util.Scanner;
public class Shift {
public static void main(String[] args) {
Scanner input = new Scanner (System.in);
int array[] = new int [5];
int array1[] = new int [5];
int i, temp;
for (i=0; i<5; i++) {
System.out.printf("Enter array[%d]: \n", i);
array[i] = input.nextInt(); //Taking input in the array
}
System.out.println("\nEntered datas are: \n");
for (i=0; i<5; i++) {
System.out.printf("array[%d] = %d\n", i, array[i]); //This will show the data you entered (Not the shifting one)
}
temp = array[4]; //We declared the variable "temp" and put the last number of the array there...
System.out.println("\nAfter Shifting: \n");
for(i=3; i>=0; i--) {
array1[i+1] = array[i]; //New array is "array1" & Old array is "array". When array[4] then the value of array[3] will be assigned in it and this goes on..
array1[0] = temp; //Finally the value of last array which was assigned in temp goes to the first of the new array
}
for (i=0; i<5; i++) {
System.out.printf("array[%d] = %d\n", i, array1[i]);
}
input.close();
}
}
Write a Java program to create an array of 20 integers, and then implement the process of shifting the array to right for two elements.
public class NewClass3 {
public static void main (String args[]){
int a [] = {1,2,};
int temp ;
for(int i = 0; i<a.length -1; i++){
temp = a[i];
a[i] = a[i+1];
a[i+1] = temp;
}
for(int p : a)
System.out.print(p);
}
}
Hello I am having difficulty implementing a counting sort method in java. I believe the problem comes from the last two loops I have in the method. I am getting an ArrayIndexOutOfBounds exception : 8. I believe this comes from my second to last for loop when at index 5 the value is 8 but I am not sure how to resolve this. Any help is appreciated. Thank you!
In my code k is the highest value in the input array.
Code:
public static void main(String[] args) {
int [] arrayOne = {0,1,1,3,4,5,3,0};
int [] output = Arrays.copyOf(arrayOne, arrayOne.length);
System.out.println(Arrays.toString(arrayOne));
countingSort(arrayOne, output, 5);
System.out.println(Arrays.toString(output));
}
public static void countingSort(int[] input, int[] output , int k){
int [] temp = Arrays.copyOf(input, k+1);
for (int i = 0; i <= k; i++){
temp[i] = 0;
}
for (int j = 0; j <= input.length - 1; j++){
temp[input[j]] = temp[input[j]] + 1;
}
for (int i = 1; i <= k; i++){
temp[i] = temp[i] + temp[i-1];
}
for (int j = input.length; j >= 1; j--){
output[temp[input[j]]] = input[j];
temp[input[j]] = temp[input[j]] - 1;
}
}
The problem is in the first loop because the array temp lenght is 6 and you are doing 7 interations in there.
So at the end of the for it is trying to do temp[6]=0 and the last position of your array is temp[5].
To fix this change your first loop to:
for (int i = 0; i < k; i++){
In the last loop you will get the same exception cause input[8] doesn't exist.
import java.util.Arrays;
public class CountingSort {
public static void main(String[] args) {
int[] input = {0,1,1,3,4,5,3,0};
int[] output = new int[input.length];
int k = 5; // k is the largest number in the input array
System.out.println("before sorting:");
System.out.println(Arrays.toString(input));
output = countingSort(input, output, k);
System.out.println("after sorting:");
System.out.println(Arrays.toString(output));
}
public static int[] countingSort(int[] input, int[] output, int k) {
int counter[] = new int[k + 1];
for (int i : input) { counter[i]++; }
int ndx = 0;
for (int i = 0; i < counter.length; i++) {
while (0 < counter[i]) {
output[ndx++] = i;
counter[i]--;
}
}
return output;
}
}
Above code is adapted from: http://www.java67.com/2017/06/counting-sort-in-java-example.html
this may help but try using the Arraya.sort() method.
e.g:
//A Java program to sort an array of integers in ascending order.
// A sample Java program to sort an array of integers
// using Arrays.sort(). It by default sorts in
// ascending order
import java.util.Arrays;
public class SortExample
{
public static void main(String[] args)
{
// Our arr contains 8 elements
int[] arr = {13, 7, 6, 45, 21, 9, 101, 102};
Arrays.sort(arr);
System.out.printf("Modified arr[] : %s",
Arrays.toString(arr));
}
}
example is a snippet from https://www.geeksforgeeks.org/arrays-sort-in-java-with-examples/
As per algorithm following implementation, I have prepared for the count sort technique
public static int[] countSort(int elements[]) {
int[] sorted = new int[elements.length+1];
int[] range = new int[getMax(elements)+1];
for(int i=0;i<range.length;i++) {
range[i] = getCount(i, elements);
try {
range[i] = range[i]+range[i-1];
}catch(ArrayIndexOutOfBoundsException ae) {
continue;
}
}
for(int i=0;i<elements.length;i++) {
sorted[range[elements[i]]] = elements[i];
range[elements[i]] = range[elements[i]]-1;
}
return sorted;
}
public static int getCount(int value,int[] elements) {
int count = 0;
for(int element:elements) {
if(element==value) count++;
}
return count;
}
public static int getMax(int elements[]) {
int max = elements[0];
for(int i=0;i<elements.length;i++) {
if(max<elements[i]) {
max = elements[i];
}
}
return max;
}
Please review and let me know if any feedback and it is more helpful.
Note :
Non-negative no won't support in the above implementation.
don't use 0th index of the sorted array.
This question already has answers here:
Java Array Sort descending?
(27 answers)
Closed 5 years ago.
This array Sorts the number in Ascending and prints the output. I wanna change it to descending, but i can't figure it out.. i tried changing the + to - but it didn't seem to work.. Can someone tip me on how to do it?
public class ArraysSorting {
public static void main(String[] args) {
int [] scores = { 5,8,2,9,3};
sortArrayDesc (scores, scores.length);
}
public static void sortArrayDesc( int [] list, int listLength) {
int index;
int smallestIndex;
int minIndex;
int temp;
for (index = 0; index < listLength - 1; index++) {
smallestIndex = index;
for (minIndex = index + 1;
minIndex < listLength; minIndex++)
if (list[minIndex] < list[smallestIndex])
smallestIndex = minIndex;
temp = list[smallestIndex];
list[smallestIndex] = list[index];
list[index] = temp;
}
//display data after sorting
System.out.print("sorted array");
for (index=0; index<listLength; index++) {
System.out.print(list[index] + ",");
}
}
}
I suggest you look up articles on "bubble sort", which is fairly close to what you are doing here. Then, follow up with learning how other sort algorithms work. You generally do not want to use an n-squared sort algorithm.
To do the sort in the reverse order, you would want to use the smallestIndex variable to store the largest index instead. Change the name there, and use the > operator on the line if (list[minIndex] < list[smallestIndex]) to find the biggest number. You will then be storing the largest number in the first list position and the smallest in the last list position.
The updated loops would look like:
for (index = 0; index < listLength - 1; index++)
{
biggestIndex = index;
for (minIndex = index + 1; minIndex < listLength; minIndex++)
if (list[minIndex] > list[biggestIndex])
biggestIndex = minIndex;
temp = list[biggestIndex];
list[biggestIndex] = list[index];
list[index] = temp;
}
A simple way to modify this code to sort array in descending order is to change the comparison operator from smaller than (<) to larger than (>).
Your code:
if (list[minIndex] < list[smallestIndex])
Change to:
if (list[minIndex] > list[smallestIndex])
You should also change the variable naming otherwise it does not make sense.
It sounds like you want a "for loop" to run in reverse:
for (i = listLength; i>0; i--) {
System.out.print(list[i] + ",");
}
You need to start with
list[listLength]
and increment down until you get to the beginning of the list/array which is
list[0].
I just made a few modifications. Changed the variable name and the important part
if (list[maxIndex] > list[largestIndex]) // changed less than sign to greater than
public class ArraysSorting {
public static void main(String[] args) {
int [] scores = { 5,8,2,9,3};
sortArrayDesc (scores, scores.length);
}
public static void sortArrayDesc( int [] list, int listLength) {
int index;
int largestIndex;
int maxIndex;
int temp;
for (index = 0; index < listLength - 1; index++) {
largestIndex = index;
for (maxIndex = index + 1;
maxIndex < listLength; maxIndex++)
if (list[maxIndex] > list[largestIndex])
largestIndex = maxIndex;
temp = list[largestIndex];
list[largestIndex] = list[index];
list[index] = temp;
}
//display data after sorting
System.out.print("sorted array");
for (index=0; index<listLength; index++) {
System.out.print(list[index] + ",");
}
}
}
This is the code I have for my selection sort program, I want to know if there's any way of improving the code without using additional methods or classes.
public class Selection_Sort {
public static void main(String[] args) {
int arr[]={234,151,123,4,5342,76,48};
int min=0; int temp;
for(int i=0;i<=arr.length-1;i++){
min=i;
for (int k=i+1;k<arr.length;k++){
if(arr[k]<arr[i]){
temp=arr[i];
arr[i]=arr[k];
arr[k]=temp;
}
}
}
for (int j=0;j<=arr.length-1;j++)
System.out.println(arr[j]+" ");
}
}
public static void main(String[] args) {
int arr[]={234,151,123,4,5342,76,48};
int arrLength = arr.length;
for(int i=0;i<arrLength-1;i++){
int min=i;
for (int k=i+1;k<arrLength;k++){
if(arr[k]<arr[min]){
min = k;
}
}
if (i != min) {
int temp=arr[i];
arr[i]=arr[min];
arr[min]=temp;
}
}
for (int j=0;j<arrLength;j++) {
System.out.println(arr[j]+" ");
}
}
Looks like you are using the bubblesort algorithm which is very slow. If you want to improve your code, i would recommend to use an algorithm like ripplesort or quicksort.
Slight improvement should be like this :
int arrayLength = arr.length;
// Then use it in conditional statement of for loop.
So that it won't invoke length property of Array every time in loop. For small number of loops it doesn't impact much but it will help to reduce the time when loops are more or number of iteration of loop are more.
The value of the local variable min is not used
k <= arr.length-1
-->
k < arr.length
Use this
class Selection {
public static void main(String[] args) {
int arr[]={234,151,123,4,5342,76,48}; /* arr[0] to arr[n-1] is the array to sort */
int lowest, i, j;
for(i = 0 ; i < arr.length-1; i++) { /* advance the position through the entire array */
lowest = i; /* assume the min is the first element */
for(j = i+1 ; j < arr.length; j++) { /* if this element is less, then it is the new minimum */
if(arr[j] < arr[lowest]) {
lowest = j; /* found new minimum; remember its index */
}
}
if(lowest != i) { /* lowest is the index of the minimum element. Swap it with the current position */
int temp = arr[i];
arr[i] = arr[lowest];
arr[lowest] = temp;
}
}
for (int k = 0; k <= arr.length-1 ; k++) {
System.out.println(arr[k] + " ");
}
}
}
This is the selection sort algorithm you asked.
Here is original Selection sort implementation. The implementation in question in not using min to perform the swap operation.
public static void sort(int[] arr) {
int min=-1;
for (int i = 0; i < arr.length; i++) {
min = i;
for (int j = i + 1; j < arr.length; j++) {
if (arr[min] > arr[j]) {
min = j;
}
}
if (min != i) {
int temp = arr[min];
arr[min] = arr[i];
arr[i] = temp;
}
}
}
public class JavaApplication55 {
public static void main(String[] args) {
int[] array ={234,435,567,768,123,456,789,789,5670,6789};
for(int j =0;j< array.length;j++){
for(int i =j+1;i < array.length;i++ ){
int temp;
if(array[j]>array[i]){
temp =array[j];
array[j] =array[i];
array[i] =temp;
}
else{}
}}
for(int k =0;k< array.length;k++){
System.out.println(array[k]);
}
}
enter code here
}