2021-07-28 数组中的逆序对

很巧的一道题，中午刚刚复习了排序算法
使用归并排序的思想

另一种方法就是暴力for循环
public class Solution {
int count =0;
public int InversePairs(int [] array) {

    merge(array,0,array.length-1);

    return count;
}
public void merge(int[] arg,int left,int right){
    if(left>=right) return;
    int mid = left+((right-left)>>1);//防止溢出
    merge(arg,left,mid);
    merge(arg,mid+1,right);
    mergeSort(arg,left,mid,right);
}
public void mergeSort(int[] arg,int left,int mid,int right){
    int p1 = mid,p2=right;
    int p=right-left;
    int[] result = new int[right-left+1];
    while(p1>=left&&p2>mid){
        if(arg[p1]>arg[p2]){
            result[p--] = arg[p1];
            count+=p2-mid;
            if(count>=1000000007){
                count%=1000000007;
            }
            p1--;

        }else{
            result[p--] = arg[p2];
            p2--;
        }

    }
    for(;p1>=left;p1--){
        result[p--] = arg[p1];
    }
    for(;p2>mid;p2--){
        result[p--] = arg[p2];
    }
    for(int i = 0;i

}