三种语言实现计算逆序对的数量（C++/Python/Java)

题目

给定一个长度为 𝑛 的整数数列，请你计算数列中的逆序对的数量。

逆序对的定义如下：对于数列的第 𝑖 个和第 𝑗 个元素，如果满足𝑖<𝑗 且𝑎[𝑖]>𝑎[𝑗]，则其为一个逆序对；否则不是。

输入格式

第一行包含整数 𝑛，表示数列的长度。

第二行包含 𝑛 个整数，表示整个数列。

输出格式

输出一个整数，表示逆序对的个数。

数据范围

1≤𝑛≤100000，
数列中的元素的取值范围 [1,109]

输入样例：

6
2 3 4 5 6 1

输出样例：

5

C++

#include <bits/stdc++.h>
#define int long long
 
using namespace std;
 
typedef long long LL;
 
 
const int N = 100010;
int a[N];
int tmp[N];
int Count = 0;
 
int merge_sort(int l, int r)
{
    if(l >= r)  return 0;
    int mid = (l + r) >> 1;
    int res = merge_sort(l, mid) + merge_sort(mid+1, r);
    int i = l, j = mid + 1;
    int k = l;
    while(i <= mid && j <= r)
    {
        if(a[i] <= a[j])    tmp[k++] = a[i++];
        else
        {
            tmp[k++] = a[j++];
            res += mid - i + 1;
        }
    }
    while(i <= mid) tmp[k++] = a[i++];
    while(j <= r)   tmp[k++] = a[j++];
    for(int t = l; t <= r; t++)
        a[t] = tmp[t];
    return res;
}
 
signed main()
{
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> a[i];
    cout << merge_sort(0,n-1) << endl;
}

Python

n = int(input())
a = [int(x) for x in input().split()]
 
tmp = [i for i in range(n)]
 
 
def merge_sort(l, r):
    if l >= r:
        return 0
    mid = (l + r) >> 1
    res = merge_sort(l, mid) + merge_sort(mid+1, r)
    i, j = l, mid + 1
    k = l
    while i <= mid and j <= r:
        if a[i] <= a[j]:
            tmp[k] = a[i]
            i += 1
        else:
            tmp[k] = a[j]
            j += 1
            res += mid - i + 1
        k += 1
    
    while i <= mid:
        tmp[k] = a[i]
        i += 1
        k += 1
    
    while j <= r:
        tmp[k] = a[j]
        j += 1
        k += 1
    
    for t in range(l ,r+1):
        a[t] = tmp[t]
        
    return res
 
 
print(merge_sort(0, n-1))

Java

import java.util.*;
 
 
public class Main{
    static public void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        int [] tmp = new int[n];
        for(int i = 0; i < n; i++)
            a[i] = sc.nextInt();
        
        System.out.print(merge_sort(a, tmp, 0, n-1));
    }
    static public long merge_sort(int[] a, int[] tmp, int l, int r){
        if(l >= r)  return 0;
    
        int mid = (l + r) >> 1;
        long res = merge_sort(a, tmp, l, mid) + merge_sort(a, tmp, mid+1, r);
        int i = l, j = mid + 1;
        
        int k = l;
        while(i <= mid && j <= r){
            if(a[i] <= a[j])    tmp[k++] = a[i++];
            else{
                tmp[k++] = a[j++];
                res += mid - i + 1;
            }
        }
        while(i <= mid) tmp[k++] = a[i++];
        while(j <= r)   tmp[k++] = a[j++];
        for(int t = l; t <= r; t++)
            a[t] = tmp[t]; 
        return res;
    }
}