35.数组中的逆序对

public class Solution {
    public int InversePairs(int [] array) {
        if(array==null || array.length==0){
            return 0;
        }
        //复制一份到array数组
        int length=array.length;
        int[] copy=new int[length];
        for(int i=0;i<length;i++){
            copy[i]=array[i];
        }
        
        int count=InversePairsCore(array,copy,0,length-1);
        return count;
    }
    
    private int InversePairsCore(int[] array,int[] copy,int low,int high){
        if(low==high){
            return 0;
        }
        int mid=(low+high)>>1;
        int leftCount=InversePairsCore(array,copy,low,mid)%1000000007;
        int rightCount=InversePairsCore(array,copy,mid+1,high)%1000000007;
        int count=0;
        //i初始化为前半段最后一个数字的下标
        int i=mid;
        //j初始化为后半段最后一个数字的下标
        int j=high;
        int locCopy=high;
        while(i>=low && j>mid){
            if(array[i]>array[j]){
                count+=j-mid;
                copy[locCopy--]=array[i--];
                if(count>=1000000007){//数值过大求余
                    count%=1000000007;
                }
            }else{
                copy[locCopy--]=array[j--];
            }
        }
        for(;i>=low;i--){
            copy[locCopy--]=array[i];
        }
        for(;j>mid;j--){
            copy[locCopy--]=array[j];
        }
        for(int s=low;s<=high;s++){
            array[s]=copy[s];
        }
        return (leftCount+rightCount+count)%1000000007;
    }
}