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?
316 Answers
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: (java.util.List,%20int)
0Just 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.
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];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:
1In 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]; } }Using array CopyGeneric 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); }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); }
}
