归并排序

链接：https://www.nowcoder.com/questionTerminal/96bd6684e04a44eb80e6a68efc0ec6c5 来源：牛客网

public class Solution {

    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;

    }

}