剑指offer 数组中的逆序对

1,2,3,4,5,6,7,0

7

public class Solution {
    private long count;
    public int InversePairs(int [] array) {
        if(array==null||array.length==0)
            return 0;
        int[] temp=new int[array.length];
        divid(array,temp,0,array.length-1);
        return (int)(count%1000000007);
    }
    public void divid(int[]array,int []temp,int low,int high){
        if(low>=high)
            return;
        int mid=(low+high)/2;
        divid(array,temp,low,mid);
        divid(array,temp,mid+1,high);
        merge(array,temp,low,high);
    }
    public void merge(int[]array,int[]temp,int low,int high){
        int lowend=(low+high)/2;
        int highpos=lowend+1;
        int temppos=low;
        int len=high-low+1;
        while(low<=lowend&&highpos<=high){
            if(array[low]<=array[highpos]){
                temp[temppos++]=array[low++];			//简化写法，节省代码
            }else{
                count+=(lowend-low+1)%1000000007;  
//改为count=count%1000000007+((middle-leftPoint+1)%1000000007);才通过
                temp[temppos++]=array[highpos++];         //简化写法，节省代码
            }
        }
        while(low<=lowend){
            temp[temppos++]=array[low++];
        }
        while(highpos<=high){
            temp[temppos++]=array[highpos++];
        }
        for(int i=len;i>0;i--,high--){					//for循环中一个分号可以写多个条件
            array[high]=temp[high];
        }
    }
}