题目描述

利用归并排序的思路，三部曲
结果用long接收，可能爆int

样例

import java.io.*;
import java.util.Scanner;
/* 2 3 4 1 5 6
三部曲：
1.全在左边
2.全在右边
3.一个在左，一个在右
*/
public class Main {
    private static long mergesort(int[] a, int left, int right) {
        if (left >= right) {
            return 0;
        }
        int mid = (right + left) / 2;
        //全在左边 + 全在右边
        long res = mergesort(a, left, mid) + mergesort(a, mid + 1, right);
        int i = left;
        int j = mid + 1;
        int[] temp = new int[right - left + 1];
        int k = 0;
        //一个在左，一个在右
        while (i <= mid && j <= right) {
            if (a[i] <= a[j]) {
                temp[k++] = a[i++];//不是逆序对就继续存放
            } else {
                temp[k++] = a[j++];//存放步骤不变
                //i右边都是>=本身,所以从大小上都符合逆序对的定义
                res += mid - i + 1;//统计逆序对的数量 
            }
        }
        while (i <= mid) {
            temp[k++] = a[i++];
        }
        while (j <= right) {
            temp[k++] = a[j++];
        }
        for (i = left, j = 0; i <= right; i++, j++) {
            a[i] = temp[j];
        }
        return res;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        int n = in.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
        }
        System.out.print(mergesort(arr,0,n-1));
    }
}