Java Tutorial
One element from the list of elements is consumed in each iteration to find its correct position.
i'th iteration we "insert" the i'th element array[i] into it's correct position among array[1], array[2], ... array[i-1] which were previously placed in sorted order.
Insertion sorting time complexity is O(N^2). here N is the number of elements in the list.
This java program is used to sort the array of integers using insertion sorting technique.
import java.util.Scanner; public class InsertionSortingTest { public static void insertionSort(int[] listVals){ int numberOfElements = listVals.length; for(int i=0;i<numberOfElements;i++){ // temp stores current element whose left side is traversed // to find correct position int temp = listVals[i]; int j = i; // checks whether left side elements are less than current element. while(j > 0 && temp < listVals[j-1]){ listVals[j] = listVals[j-1]; j = j - 1; } // moving current element to correct position in left side. listVals[j] = temp; } } public static void main(String[] args){ int[] numberList; Scanner in = new Scanner(System.in); System.out.println("Insertion Sorting"); System.out.println("--------------------"); System.out.println("Enter numbers count: "); int count = in.nextInt(); numberList = new int[count]; System.out.println("Enter numbers: "); for (int i=0;i<count;i++){ numberList[i]=in.nextInt(); } in.close(); insertionSort(numberList); System.out.println("Sorted numbers: "); for(int number : numberList){ System.out.print(number+"\t"); } } }Output:
$ javac InsertionSortingTest.java $java InsertionSortingTest Insertion Sorting -------------------- Enter numbers count: 5 Enter numbers: 25 18 19 35 16 Sorted numbers: 16 18 19 25 35
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