题解 | #数组中的逆序对#

public class Solution {

    private int count =0;
      int [] temp=new int[100005];

    public void mergrSort(int [] a,int [] tem,int l,int r)
    {
        if(l<r)
        {
             int mid=l+(r-l)/2;;
// System.out.println(mid);
             mergrSort(a,tem,l,mid);
             mergrSort(a,tem,mid+1,r);
          int l1=l,r1=mid+1,k=0;
          while(l1<=mid&&r1<=r)
          {
             if(a[l1]<=a[r1])
             {
                 tem[k++]=a[l1++];
             }
             else {


                 tem[k++]=a[r1++];
                 count+=(mid-l1+1);
                 count=count%1000000007;
             }

          }
          while(l1<=mid)
          tem[k++]=a[l1++];
          while(r1<=r)
          {
              tem[k++]=a[r1++];
          }
          for(int i=0;i<k;++i)
          a[l+i]=tem[i];

        }

    }
    public int InversePairs(int [] array) {
        int n=array.length;
        if(n<=1)
        return 0;
      
        
        mergrSort(array,temp,0,n-1);
        

        return count;


    }
}

使用归并排序，