二维数组的最大联通子数组

·主要思路：二维连通数组求最大子数组，我们在这里主要运用了降维的思想，主要是通过将二维数组转化为一维数组求最大子数组的思想，在项目中定义一个算法，用于实现一行连续数组的最大子数组，并用p q记录最大子数组的起始结束位置下标。在主函数中利用for循环实现二维数组每一行最大子数组的求和，在利用循环实现最大子数组的连接，添加对单独正数据的判断，从而实现连通的最大子数组问题。
·代码如下

import java.util.Scanner;
 
public class Test
{
    static int q=0,p=0;
    static Scanner str=new Scanner(System.in);
    public static void main(String args[])
    {
        int m1=0,m2=0;  
        int max=0;
        int sum=0;
        System.out.println("输入二维数组的行列数");
        m1=str.nextInt();
        m2=str.nextInt();
        int [][]a=new int[m1][m2];
        int []b=new int[m2];
        int []left=new int[m2];
        int []right=new int[m2];
        int []t=new int[m2];
        for (int i = 0; i < m1; i++)
        {
            for (int j = 0; j < m2; j++)
            {
                a[i][j]=str.nextInt();  
            }
        }
         for (int i = 0; i<m1; i++)//求每一行最大子数组
         {
             for (int j = 0; j<m2; j++)  
             {      
                 b[j] = a[i][j];   
             } 
             sum = findmax(m1, b, p, q);    
             left[i] = p;     //记录最大子数组的坐标位置
             right[i] = q;    
             t[i] = sum;
         }
            max = t[0];
            for (int i = 0; i + 1<m2; i++)//将最大子数组合并
            {
                if (left[i] <= right[i + 1] && right[i] >= left[i + 1])//两行的最大子数组块相连
                {
                    max += t[i + 1];
                }
                for (int j = left[i]; j<left[i + 1]; j++)
                {
                    if (a[i + 1][j]>0)
                        max += a[i + 1][j];                   //判别独立正数
                }
            }
         System.out.println("最大子数组和为："+max); 
    }   
    public static int findmax(int n,int a[],int p, int q )
    {
        int []b=new int[a.length+1]; 
        b[a.length]=0;
        int sum1=0;
        int max1=0;
        for(int i=0;i<n;i++)
        {
            if(sum1<0)
            {
                sum1=a[i];
            }
            else
            {
                sum1=sum1+a[i];
            }
            b[i] = sum1;
        }
        max1=b[0];
        for(int i=0;i<n;i++)
        {
            if(max1<b[i])
            {
                max1=b[i];
                q=i;
            }
        }
        for(int i=q;i>=0;i--)
        {
            if(b[i]==a[i])
            {
                p=i;
                break;
            }
        }
        return max1;
    }   
}

·结果截图：