AcWing 788. 逆序对的数量（归并排序应用）

题目链接：https://www.acwing.com/problem/content/790/

解法：
主要的问题集中在第22行的处理，记住一个原则：谁流失，统计谁！一旦j++，那么arr[j]将不可能被统计到，所以应该统计所有大于arr[j]的右边的数

AC代码：

import java.util.*;

public class Main {
    static int N = (int) 1e5 + 10;
    static int[] arr = new int[N], t = new int[N];
    
    
    static long mergeCount(int[] arr, int l, int r) {
        if (l >= r) return 0;
        
        long res = 0;
        int mid = l + r >> 1;
        
        res += mergeCount(arr, l, mid);
        res += mergeCount(arr, mid + 1, r);
        
        int i = l, j = mid + 1;
        int pos = 0;
        while (i <= mid && j <= r) {
            if (arr[i] <= arr[j]) t[pos ++] = arr[i ++];
            else {
                res += mid - i + 1;
                t[pos ++] = arr[j ++];
            }
        }
        
        
        while (i <= mid) {
            t[pos ++] = arr[i ++];
        }
        
        while (j <= r) {
            t[pos ++] = arr[j ++];
        }
        
        for (int k = 0; k < pos; k ++) arr[l + k] = t[k];
        
        return res;
    }
    
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        
        for (int i = 0; i < n; i ++) arr[i] = sc.nextInt();
        
        long res = mergeCount(arr, 0, n - 1);
        
        System.out.println(res);
    }
}