数组中的逆序对

题目：在数组中的两个数字，如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组,求出这个数组中的逆序对的总数P。并将P对1000000007取模的结果输出。 即输出P%1000000007 

思路：基于归并排序的思路

 public int InversePairs(int [] array) {
        if(array==null||array.length==0)
        {
            return 0;
        }
        int[] copy = new int[array.length];
        for(int i=0;i<array.length;i++)
        {
            copy[i] = array[i];
        }
        int count = InversePairsCore(array,copy,0,array.length-1);//数值过大求余
        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;
        int i=mid;
        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;
    }