一、题目内容

二、题解

解法1：

思路

count+=mid+1-i;
count%=1000000007;

代码

public class Solution {
    int count = 0;
    public int InversePairs(int [] array) {
        if(array.length<2){
            return 0;
        }
        mergeSort(array,0,array.length-1);
        return count;
    }
    private void mergeSort(int[] array,int left,int right){
        int mid = left+(right-left)/2;
        if(left<right){
            // 左子数组
            mergeSort(array,left,mid);
            // 右子数组
            mergeSort(array,mid+1,right);
            merge(array,left,mid,right);
        }
    }
    private void merge(int[] array,int left,int mid,int right){
        int[] arr = new int[right-left+1];
        int arrIdx = 0;
        int i = left,j = mid+1;
        while(i<=mid&&j<=right){
            if(array[i]<=array[j]){
                arr[arrIdx++] = array[i++];
            }else{
                arr[arrIdx++] = array[j++];
                count+=mid+1-i;
                count%=1000000007;
            }
        }
        while(i<=mid){
            arr[arrIdx++] = array[i++];
        }
        while(j<=right){
            arr[arrIdx++] = array[j++];
        }
        for(int num:arr){
            array[left++] = num;
        }
    }
}