[劍指offer] 數組中的逆序對 java （歸併排序）

public class Solution {
    int p;
    public int InversePairs(int [] array) {
        p=0;
        if(array!=null){
            mergesort(array,0,array.length-1);                
        }
        return p;
    }
    
    public void mergesort(int[] arr,int l,int r){
        if(l>=r) return;
        int mid=(l+r)/2;
        mergesort(arr,l,mid);
        mergesort(arr,mid+1,r);
        merge(arr,l,mid,r);
    }
  
    public void merge(int[] arr,int l,int mid,int r){
        int[] help=new int[r-l+1];
        int p1=l,p2=mid+1;
        int i=0;
        while(p1<=mid&&p2<=r){
            if(arr[p1]<=arr[p2]){
                help[i++]=arr[p1++];
            }else{
                help[i++]=arr[p2++];
                p+=mid-p1+1;
                //後面數組的數字跳過了前面數組剩餘的數字，剩餘多少個就是多少的逆序對
                p%=1000000007;
            }
            
        }
        while(p1<=mid){
            help[i++]=arr[p1++];
        }
        while(p2<=r){
            help[i++]=arr[p2++];
        }
        for(int j=0;j<help.length;j++){
            arr[l+j]=help[j];
        }
    }
}