straight insertion sort

Insert one value into a sorted table, then get a new table which has increased one value in order. It is stable and will take O(n²).

Like the following picture shows:

The java example is:

public class StraightInsertionSort {
    public static void main(String[] args) {
        int a[] = {3,1,5,7,2,4,9,6,10,8};
        StraightInsertionSort obj=new StraightInsertionSort();
        System.out.println("Initial value：");
        obj.print(a);
        obj.insertSort(a);
        System.out.println("nAfter sort：");
        obj.print(a); 
    }
    public void print(int a[]){
        for(int i=0;i<a.length;i++){
            System.out.print(a[i]+" "); 
        } 
    }
    public void insertSort(int[] a) {
        for(int i=1;i<a.length;i++){
            int j;
            int x=a[i];//x is the value need to be inserted
            for( j=i;  j>0 && x<a[j-1];j--){
                a[j]=a[j-1]; 
            }
            a[j]=x;//insert
        } 
    }
}

The result will be:

Initial value：
3 1 5 7 2 4 9 6 10 8 
After sort：
1 2 3 4 5 6 7 8 9 10