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);
}
}
I need my array to be the size of i once iterated through the for loop. Right now it is saying that my return "array" is not found.
public int[] method(int[] a, int red, int yellow) {
for (int i = 0; i < length; i++) {
int[] array = new int[i];
array[i] = a[i];
}
As you have defined the array inside the for loop, the scope of variable array ends when the for loop ends. This is why the compiler can't find the array variable while returning. Try shifting array variable before the for loop.
You should defined the array outside of the loop. Once a variable declared inside the loop it's scoped will limited to the loop.
If you think this would make sense according to your logic you could try the below code. However you will face another issue with this IndexOutOfBoundException. I suggest to debug and work more on your logic
int[] array = null;
for (int i = 0; i < a.length; i++) {
array = new int[i];
if (a[i] >= red && a[i] <= yellow) {
array[i] = a[i];
}
}
return array;
try this.you can't return array inside the for loop as well
public static void main(String arg[]){
int[] a={1,2,3,4,5,6};
method(a,1,4);
}
public static void method(int[] a, int x, int y) {
int[] array = new int[a.length];
for (int i = 0; i < a.length; i++) {
if (a[i] >= x && a[i] <= y) {
array[i] = a[i];
}
}
System.out.println(Arrays.toString(array));
}
}
I'm trying to make a selection sort algorithm in java that finds the smallest element of an unsorted array and puts it on the end of a new array. But my program only copies the first element twice, gets the next one, and then the rest is all zeroes:
public static int find_min(int[] a){
int m = a[0];
for (int i = 0; i < a.length; i++){
if (m > a[i])
m = a[i];
}
return m;
}
public static int[] selectionSort(int[] unsorted){
int[] c = new int[unsorted.length];
for(int i = 0; i < unsorted.length; i++){
int smallest = find_min(unsorted);
c[i] = smallest;
unsorted = Arrays.copyOfRange(unsorted, i+1, unsorted.length );
}
return c;
}
When I put in something like:
public static void main(String[] args){
int a[] = {1,-24,4,-4,6,3};
System.out.println(Arrays.toString(selectionSort(a)));
}
I get:
[-24, -24, -4, 0, 0, 0]
where is this going wrong? Is this a bad algorithm?
Ok, let's revisit the selection sort algorithm.
public static void selectionSort(int[] a) {
final int n = a.length; // to save some typing.
for (int i = 0; i < n - 1; i++) {
// i is the position where the next smallest element should go.
// Everything before i is sorted, and smaller than any element in a[i:n).
// Now you find the smallest element in subarray a[i:n).
// Actually, you need the index
// of that element, because you will swap it with a[i] later.
// Otherwise we lose a[i], which is bad.
int m = i;
for (int j = m + 1; j < n; j++) {
if (a[j] < a[m]) m = j;
}
if (m != i) {
// Only swap if a[i] is not already the smallest.
int t = a[i];
a[i] = a[m];
a[m] = t;
}
}
}
In conclusion
You don't need extra space to do selection sort
Swap elements, otherwise you lose them
Remembering the loop invariant is helpful
class arrayDemo {
static void sort2D(int[][] B) {
boolean swap = true;
int oy=0;
int temp=0;
for(int ox=0;ox<B.length;ox++){
while(oy<B[ox].length) {
while(swap) {
swap = false;
for(int ix=0;ix<B.length;ix++) {
for(int iy=0;iy<B[ix].length;iy++) {
if(B[ox][oy]<B[ix][iy]) {
temp = B[ix][iy];
B[ix][iy] = B[ox][oy];
B[ox][oy] = temp;
swap = true;
}
}
}
}
oy++;
}
}
for(int row=0;row<B.length;row++)
for(int col=0;col<B[row].length;col++)
System.out.println(B[row][col]);
}
public static void main(String...S) {
int y[][] = {{10,20,0,30},{10,5,8},{3,9,8,7},{2,3}};
sort2D(y);
}
}
I am trying to sort a 2D array in ascending order.
Input: {{10,20,0,30},{10,5,8},{3,9,8,7},{2,3}};
Output: 30,20,10,10,9,8,8,7,5,3,0,2,3
Can someone help me know what is wrong with my code.
You are comparing elements that are not in the same row or column. Each sub-array should be sorted individually. You might want to reconsider this line if (B[ox][oy] < B[ix][iy]).
That code has a number of problems.
It throws ArrayIndexOutOfBoundsException. This is because all for loop tests test against B.length, which is not correct for inner arrays.
You are comparing every pair of elements, but some pairs are the reverse of other pairs, and the reverse pairs should not be tested. You need to limit the scope of your inner set of for loops, by starting at a different index.
To fix all these problems, the path of least resistance is to dump the 2D array into a 1D array and sort that, which is much easier.
Here is code that has been tested and shown to work:
static void sort2D(int[][] B) {
int count = 0;
for (int[] is : B)
for (int i : is)
count++;
int[] A = new int[count];
count = 0;
for (int[] is : B)
for (int i : is)
A[count++] = i;
int temp;
for (int i = 0; i < A.length; i++)
for (int j = i + 1; j < A.length; j++)
if (A[i] > A[j]) {
temp = A[i];
A[i] = A[j];
A[j] = temp;
}
for (int i = 0; i < A.length; i++)
System.out.print(A[i] + ",");
}
For this assignment, I am unable to get the last method to loop again and find the next largest element in the series of random numbers. Some input to solve this problem would be great. Thank you.
public static void main(String[] args) {
int[] array = randomIntArray (10);
sortArray (array);
}
public static int randomInt (int low, int high){ // Create a serie of random numbers
int x = 0;
for (int i = 0; i < 10; i++){
x = (int)(Math.random ()* (high - low) +low);
}
return x;
}
public static int[] randomIntArray (int n) { // Size of array
int[] a = new int [n];
for (int i = 0; i <a.length; i++){
a[i] = randomInt (-5, 15);
}
return a;
}
public static int indexOfMaxInRange (int[] a, int low , int high){
int index = low;
for (int i = low +1; i < high; i++){
if (a[i] > a[index]){ // If the position of i is greater than index
index = i; // then index will equal that position with the highest element
}
}
return index;
}
public static int swapElement (int []a, int index , int i){
int tmp =0;
for(i = 0; i < (a.length); i++) {
tmp = index;
a[i]= a[tmp] ;
}
return a[tmp] ;
}
public static void sortArray (int[] array){ //The sortArray calls the indexOfMaxInRange to get the index of the largest element, then uses swapElement to swap that index's position to position a[i].
for (int b= 0; b <array.length; b++){
System.out.println (array[b]+"\t"+ b); // Print out original list of random numbers
}
System.out.println ();
for (int i = 0; i <array.length; i++){
int index = indexOfMaxInRange (array, 0, 10 );
int sort = swapElement (array, index, 0);
System.out.println (sort+"\t"+ i);
}
}
This looks like homework, since sorting an array is handled by umpteen available libraries, so I am going to try to give hints rather than a full answer.
Your randomInt function does 10 times more work than it needs to. This won't change the behavior of your program but it's a waste.
Your swapElement function doesn't look at all right. You don't need to return anything, and you don't need to loop. Just swap the elements in positions "index" and "i". Maybe call the arguments "a" and "b" or "pos1" and "pos2" to be less confusing.
Inside sortArray, after you find the largest element and put it in position 0, you need to keep going and find the 2nd largest element and put it in position 1, and so forth. You will need to have a loop for this. You have the pieces almost ready. Hint: you can find the 2nd largest element, after you've put the largest one in position 0, by finding the largest remaining element (i.e. call indexOfMaxInRange but pass 1 as the start position).
Changes made to the last 3 methods
public static int indexOfMaxInRange (int[] a, int low , int high){
int index = low;
for (int i = low +1; i < a.length; i++){
if (a[i] > a[index]){ // If the position of i is greater than index
index = i; // then index will equal that position with the highest element
}
}
return index;
}
public static void swapElement (int []a, int index , int i){
int pos2;
pos2 = index;
a[i]= a[pos2];
System.out.println (a[i]+"\t"+ index);
}
public static void sortArray (int[] array){
for (int b= 0; b <array.length; b++){
System.out.println (array[b]+"\t"+ b);
}
System.out.println ("Number in order");
for (int i = 0; i <array.length-1; i++){
int index = indexOfMaxInRange (array, i, 10 );
swapElement (array, index,i );
}
}