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

public class Solution {
    public int InversePairs(int [] array) {
        if(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=inversePairsMethod(array,copy,0,array.length-1);

        return count ;
    }

    private static int inversePairsMethod(int[] array, int[] copy, int start, int end) {
        if(start == end){
            copy[start]=array[start];
            return 0;
        }
        int len = (end - start)/2;

        int left = inversePairsMethod(copy, array, start, start+len)% 1000000007;
        int right = inversePairsMethod( copy,array, start + len +1, end)% 1000000007;
        int i = start + len;
        int j = end;
        int index = end;
        int count = 0;
        while (i>=start && j >=start + len +1){
            if(array[i] > array[j]){
                copy[index--]=array[i--];
                count += j - start - len;
                if(count>=1000000007)//数值过大求余
                {
                    count%=1000000007;
                }
                //count++;
            }else {
                copy[index--] = array[j--];
            }
        }
        for(;i>=start;--i){
            copy[index--] = array[i];
        }
        for(;j>=start + len + 1;--j){
            copy[index--] = array[j];
        }
        return (count+left+right) % 1000000007;
    }
}