Insertion Sort 与 Merge Sort的性能比较(Java)

 　　 public static void main(String[] args)
     {
         Scanner input = new Scanner(System.in);
         int n = input.nextInt();
         int[] a = new int[n];

         //生成n个随机数
         for(int i = 0; i < n; i++)
             a[i] = (int)(Math.random() * 100);

         /*long start = System.currentTimeMillis();
         InsertionSort(a);
         long end = System.currentTimeMillis();*/
         //100000个测试数据InsertionSort的排序时间约为2800ms

         /*System.out.println(Arrays.toString(a));
         System.out.println(start);
         System.out.println(end);
         System.out.printf("The time is: %d", end - start);*/

         //获取当前时间(ms)
         long start = System.currentTimeMillis();
         divide(a, 0, a.length - 1);
         long end = System.currentTimeMillis();
         //100000000个测试数据MergeSort的排序时间约为12000ms 10000000个测试数据用时约为1200ms
         //相对于平均时间复杂度O(n^2)的插入排序 随着测试数据数量级的增长 性能提升极大

         //System.out.println(Arrays.toString(a));
         System.out.println(start);
         System.out.println(end);
         // end - start 即为排序算法运行的时间
         System.out.println("The time is: " + (end - start));
     }

     /*public static void InsertionSort(int[] a)
     {
         for(int i = 1; i < a.length; i++)
         {
             int key = a[i];
             int j;
             for(j = i - 1; j >= 0 && key < a[j]; j--)
                 a[j + 1] = a[j];
             a[j + 1] = key;
         }
     }*/

     public static void divide(int[] a, int front, int rear)
     {
         if(rear <= front)
             return;

         int mid = front + (rear - front) / 2;
         divide(a, mid + 1, rear);
         divide(a, front, mid);
         merge(a, front, mid, rear);
     }

     public static void merge(int[] a, int front, int mid, int rear)
     {
         int[] b = new int[rear - front + 1];
         int i = front, j = mid + 1, k = 0;

         while(i <= mid && j <= rear)
             if(a[i] < a[j])
                 b[k++] = a[i++];
             else
                 b[k++] = a[j++];

         if(i > mid)
             while(j <= rear)
                 b[k++] = a[j++];
         else
             while(i <= mid)
                 b[k++] = a[i++];

         for(int t = 0, q = front; t < b.length; t++, ++q)
             a[q] = b[t];
     }