[剑指offer]数组中的逆序对

思路：
可以看成归并排序，两两比较之后存入数组并排序，存储该逆序对的数量
实现：

public class Solution {
    public int InversePairs(int [] array) {
        if(array == null || array.length <= 0) return 0;
        int pairsNum = mergeSort( array , 0 , array.length-1 );
        return pairsNum;
    }
    public int mergeSort(int [] array ,int left , int right){
        if(left == right ) return 0;
        int mid = (left + right) / 2;
        int leftNum = mergeSort( array , left , mid );
        int rightNum = mergeSort( array , mid+1 , right );
        return ( Sort( array , left , mid , right) + leftNum + rightNum ) % 1000000007;
    }
    public int Sort(int []array , int left , int mid , int right ){
        int cur1 = mid;
        int cur2 = right;
        int cur3 = right - left;
        int temp[] = new int[right-left+1];
        int pairsnum = 0;
        while(cur1 >= left && cur2 >= mid+1 ){
            if(array[cur1] > array[cur2] ){
                temp[cur3--] = array[cur1--];
                pairsnum += (cur2 - mid);
                if(pairsnum > 1000000007){
                    pairsnum %= 1000000007;
                }
            }
            else {
                temp[cur3--] = array[cur2--];
            }
        }
        while(cur1 >= left){
            temp[cur3--] = array[cur1--];
        }
        while(cur2 >= mid + 1){
            temp[cur3--] = array[cur2--];
        }
        int i = 0;
        while(left <= right){
            array[left++] = temp[i++];
        }
        return pairsnum;
    }
}