Question

我想找到2D数组中可能的反转次数。我编写了这个程序，需要加快速度的方法：

import java.io.*;
import java.util.*;
class Solution
{
static long merge(int[] array, int[] left, int[] right)
{
    int i = 0, j = 0, count = 0;
    while (i < left.length || j < right.length)
    {
        if (i == left.length)
        {
            array[i+j] = right[j];
            j++;
        }
        else if (j == right.length)
        {
            array[i+j] = left[i];
            i++;
        }
        else if (left[i] <= right[j])
        {
            array[i+j] = left[i];
            i++;
        }
        else
        {
            array[i+j] = right[j];
            count += left.length-i;
            j++;
        }
    }
    return count;
}

static long invCount(int[] array)
{
    if (array.length < 2)
        return 0;

    int m = (array.length + 1) / 2;
    int left[] = Arrays.copyOfRange(array, 0, m);
    int right[] = Arrays.copyOfRange(array, m, array.length);

    return invCount(left) + invCount(right) + merge(array, left, right);
}

public static void main(String args[])
{
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    long inversions=0;
    int[][] arr2=new int[n][n];
    if(n<0)
    {
        System.out.println("0");
        return;
    }
    for (int i=0;i<n;i++)
        for (int j=0;j<n;j++)
        {
            arr2[i][j]=sc.nextInt();
            //arr2[i][j]=arr[i][j];
        }

    long inv=0;
    int counter=0;
    for(int i=0;i<n-1;i++)
    {
        for(int j=i+1;j<n;j++)
        {
            while(counter<n)
            {
                for(int z=counter;z<n;z++)
                {
                    //System.out.println("comparing "+arr2[i][counter]+" with "+arr2[j][z]);
                    if(arr2[i][counter]>arr2[j][z])
                        inv++;
                }
                counter++;
                //System.out.println("end while---------\n");
            }
            //System.out.println("Row change#########\n");
            counter=0;
        }
    }
    for (int i=0;i<n;i++)
      inv=inv+invCount(arr2[i]);
    System.out.println(inv);
}
}

此程序可以优化吗？或者这个程序在某处错了？我得到了2个测试用例的正确输出：2D阵列的4次反转：

9 7
1 2

和19个二维数组反转：

9 7 6
1 2 5
2 3 1

感谢您的帮助。:)

Answer 1

                inversionCount=0;
                for(R=0;R<sizeOf2DArray;R++) {
                    for(C=0;C<sizeOf2DArray;C++) {
                        firstElem=inputArray[R][C];
                        for(int j=R;j<sizeOf2DArray;j++) {
                            for(int i=C;i<sizeOf2DArray;i++) {
                                secondElem=inputArray[j][i];
                                if(firstElem>secondElem) {
                                    inversionCount++;
                                }
                            }
                        }
                    }   
                }
                System.out.println(inversionCount);

计算2d数组的反转

1 个答案: