查找数组连成环形的和最大的连续子数组

思路：

　　把每一种起点情况下的最大子数组之和S求出，存入S[]数组中，最后比较S[]中的最大值（i为数组的长度）存为MaxSum。而此时的起点-finalStart和终点-finalEnd也同样可以在求MaxSum的同时记录下来。

代码如下：

package Test;
 
import java.util.Scanner;
 
 
public class MaxSubArray {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
         
         

        System.out.println("输入数组长度");
        int n = scan.nextInt();
        int[] a = new int[n];
         
        System.out.println("输入数组元素");
        for(int i = 0;i < n;i++)
        {
            a[i] = scan.nextInt();
        }
        scan.close();
    
        int[] result = maxSub(a,a.length);
        System.out.println("不连接成环的和最大的连续子数组：");
        for(int i = result[0];i <= result[1];i++)
        {
            System.out.print(a[i] + "	");
        }
        System.out.println("和为：" + result[2]);
     
     
         
         
       
      
        int[] b = new int[2 * n - 1];
        for(int i = 0;i < n - 1;i++)
        {
            b[i] = a[i];
            b[n + i] = a[i];
        }
        b[n - 1] = a[n - 1];
        int[] result2 = maxSub(b,n);
        System.out.println("

将数组连成环后的和最大的连续子数组：");
        for(int i = result2[0];i <= result2[1];i++)
        {
            System.out.print(b[i] + "	");
        }
        System.out.println("和为：" + result2[2]);
 
         
    }
     
     
     
     
  
    public static int[] maxSub(int[] a,int n)
    {
        int an = a.length;
        int currectSum = a[0];
        int currectStartIndex = 0;
        int count = 1;
        int[] result = new int[3];
        result[0] = 0;
        result[1] = 0;
        result[2] = a[0];
        for(int i = 1;i < an;i++)
        {
            if(currectSum <= 0)
            {
                currectSum = a[i];
                currectStartIndex = i;
                count = 1;
            }
            else
            {
                currectSum += a[i];
                count++;
            }
            if(currectSum > result[2])
            {
                result[0] = currectStartIndex;
                result[1] = i;
                result[2] = currectSum;
            }
            if(count >= n)
            {
                break;
            }
        }
        return result;
    }
     
     
     
}