LeetCode algorithm problem: an array of reverse reversePairs

Two numbers in the array, if a number is greater than the front behind the figures, the two numbers form a reverse pair. Enter an array, this array P. finds the total number of reverse pair P and outputs the result modulo 1000000007. I.e., the output P% 1000000007
Input Description:
Title ensure the same number of the input array is not

data range:

对于%50的数据,size<=10^4

对于%75的数据,size<=10^5

对于%100的数据,size<=2*10^5

Recursive sequencing method

class Solution {  
    private int[] aux;
    private int p;
    public int reversePairs(int [] array) {
        int len;
        if(array == null || (len=array.length) == 0)return 0;
        aux = new int[len];
        mergeSort(array,0,len - 1);
        return p;
    }
    private void mergeSort(int [] array,int l,int h) {
        if(h <= l)return;
        int m = l + (h-l) / 2;
        mergeSort(array,l,m);
        mergeSort(array,m + 1,h);
        merge(array,l,m,h);
    }
    
    private void merge(int [] array,int l,int m,int h) {
        int i = m, j = h;
        for(int k = l;k <= h;k++) 
            aux[k] = array[k];
        for(int k = h;k >= l;k--) {
            if(i < l) {
                array[k] = aux[j--];
            }else if(j <= m) {
                array[k] = aux[i--];
            }else if(aux[i] > aux[j]) {
                p += j - m;
                array[k] = aux[i--];
            }else {
                array[k] = aux[j--];
            }
        }
    }
}