Merge sort and solve minor problems and to reverse the problem

Small and issues:

The left than the element's current small elements and

Violent solution is not recommended O (N ^ 2)

Merge acceleration

    public static  void  main(String[] args){
        int [] a=new int[]{2, 4, 1, 3, 5};
        System.out.println( smallSum(a));
    }

    public static int smallSum(int[] arr) {
        if (arr == null || arr.length < 2) {
            return 0;
        }
        return mergeSort(arr, 0, arr.length - 1);
    }

    public static int mergeSort(int[] arr, int l, int r) {
        if (l == r) {
            return 0;
        }
        int mid = l + ((r - l) >> 1);
        return mergeSort(arr, l, mid) + mergeSort(arr, mid + 1, r) + merge(arr, l, mid, r);
    }

    public static int merge(int[] arr, int l, int m, int r) {
        int[] help = new int[r - l + 1];
        int i = 0;
        int p1 = l;
        int p2 = m + 1;
        int res = 0;
        while (p1 <= m && p2 <= r) {
            //res += arr[p1] < arr[p2] ? (r - p2 + 1) * arr[p1] : 0;//小和
            res += arr[p1] > arr[p2] ? (m-p1+1)  : 0;//逆序
            help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
        }
        while (p1 <= m) {
            help[i++] = arr[p1++];
        }
        while (p2 <= r) {
            help[i++] = arr[p2++];
        }
        for (i = 0; i < help.length; i++) {
            arr[l + i] = help[i];
        }
        return res;
    }