挑战程序设计竞赛（算法和数据结构）——7.6逆序数的JAVA实现

Posted 2021-12-15 小乖乖的臭坏坏

逆序数的题简单来说就可以归化成判断归并排序需要交换几次的问题，因此代码可以直接沿用归并排序。
思路讲解：

import java.io.BufferedInputStream;
import java.util.Scanner;

public class InversionNumber 
    public static void main(String[] args) 
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        int n = cin.nextInt();
        int[] S = new int[n];
        for (int i=0;i<n;i++)
            S[i] = cin.nextInt();
        
        int cnt = mergeSort(S, 0, n);
        System.out.println(cnt);
    

    public static int merge(int[] A, int left, int mid, int right)
        int n1 = mid - left;
        int n2 = right - mid;
        int cnt = 0;
        int [] L = new int[n1+1];
        int [] R = new int[n2+1];
        for(int i=0;i<n1;i++)
            L[i] = A[left+i];
        
        for(int j=0;j<n2;j++)
            R[j] = A[mid+j];
        
        L[n1] = Integer.MAX_VALUE;
        R[n2] = Integer.MAX_VALUE;
        int i = 0;
        int j = 0;
        for(int k = left;k<right;k++)
            if(L[i]<=R[j])
                A[k] = L[i];
                i++;
            
            else
                A[k] = R[j];
                j++;
                cnt+=n1-i;
            
        
        return cnt;
    

    public static int mergeSort(int[] A, int left, int right)
        if(left+1<right)
            int mid = (left+right)/2;
            int v1,v2,v3;
            v1 = mergeSort(A,left, mid);
            v2 = mergeSort(A,mid,right);
            v3 = merge(A, left, mid, right);
            return v1+v2+v3;
        
        else return 0;
    

输入：

5
3 5 2 1 4

输出：

6