最大範圍和編碼評估和演示挑戰

Question

我試圖解決最大距離和這是一個代碼的eval挑戰這是這樣的：

鮑勃正在制定新的戰略，以獲得豐富的股市。他希望將自己的投資組合投資1天或更長時間，然後在合適的時間出售以最大限度地提高收益。鮑勃苦心追蹤了他的投資組合在過去N天裏每增加或減少多少。現在他已經僱傭你去弄清楚他的投資組合能夠取得的最大收益。

例如：

鮑勃跟蹤股票市場的最後10天。上的每一天中，增益/損失如下：

7 -3 -10 4 2 8 -2 4 -5 -2

如果Bob進入第4天股市和退出在第8天（ 5天計），他的收益將一直

16（4 + 2 + 8 + 2 + 4）

我的輸入文件包括：

5;7 -3 -10 4 2 8 -2 4 -5 -2 
6;-4 3 -10 5 3 -7 -3 7 -6 3 
3;-7 0 -45 34 -24 7 

我索姆ehow得到這三條線的輸出爲零。 我試過調試，但沒有得到該問題。 這裏是我的代碼：

package MaxRangeSum; 

import java.io.BufferedReader; 
import java.io.FileNotFoundException; 
import java.io.FileReader; 
import java.io.IOException; 
import static java.lang.Math.floor; 
import java.util.StringTokenizer; 

/** 
* 
* @author kanua_000 
*/ 
public class MaxRangeSum { 

    public static int ret_cross = 0,ret_sum = 0; 
    public static Integer leftsum = Integer.MIN_VALUE; 
    public static Integer rightsum = Integer.MIN_VALUE; 
    public static int sum_cross = 0, maxleft = 0,sum_max = 0; 
    public static int leftsum_subarray = 0; 
    public static int rightsum_subarray = 0; 
    public static int crosssum = 0; 


    public static int max_crossing_subarray(int[] A, int low, int mid, int high) { 


     int i = mid; 
     while (i != low) { 
      sum_cross = sum_cross + A[i]; 
      if (sum_cross > leftsum) { 
       leftsum = sum_cross; 
       // maxleft = i; 
      } 
      --i; 
     } 


     // int maxright = 0; 
     sum_cross = 0; 
     int j = mid + 1; 

     while (j != high) { 
      sum_cross = sum_cross + A[j]; 
      if (sum_cross > rightsum) { 
       rightsum = sum_cross; 
      // maxright = j; 
      } 
      ++j; 
     } 

     ret_cross = leftsum + rightsum; 

     return (ret_cross); 

    } 

    public static int max_subarray(int[] a, int low, int high) { 

     int i = 0,j = 0; 
     int[] A = new int[50]; 

     if (low == high) { 
      return (A[low]); 
     } else { 

      int mid = (int) (floor((low + high)/2)); 

      while (i != mid) { 
       ret_sum = ret_sum + A[i]; 
       if (ret_sum > leftsum_subarray) { 
        leftsum_subarray = ret_sum; 
       } 
       i++; 
      } 

      leftsum_subarray = max_subarray(A, low, mid); 

      j = mid+1; 

      while (j != high) { 
       ret_sum = ret_sum + A[j]; 
       if (ret_sum > rightsum_subarray) { 
        rightsum_subarray = ret_sum; 
        // maxleft = i; 
       } 
       j++; 
      } 

      rightsum_subarray = max_subarray(A, (mid + 1), high); 



      crosssum = max_crossing_subarray(A, low, mid, high); 

      if ((leftsum_subarray >= rightsum_subarray) & (leftsum_subarray >= crosssum)) { 
       return (leftsum_subarray); 
      } else if ((rightsum_subarray >= leftsum_subarray) & (rightsum_subarray >= crosssum)) { 
       return (rightsum_subarray); 
      } else { 
       return (crosssum); 
      } 
     } 

    } 

    public static void main(String args[]) throws FileNotFoundException, IOException { 

     int n = 0, i = 0; 

     int line1[] = new int[50]; 
     int line2[] = new int[10]; 
     int line3[] = new int[10]; 
     int line4[] = new int[6]; 

     try (BufferedReader in = new BufferedReader(new FileReader("E:\\FresnoState\\CSCI 174\\MaxRangeSum.txt"))) { 

      String line; 
      while ((line = in.readLine()) != null) { 
       StringTokenizer tokenizer = new StringTokenizer(line, " "); 

       while (tokenizer.hasMoreTokens()) { 

        String digits = tokenizer.nextToken(); 

        if (digits.contains(";")) { 
         digits = (digits.substring(2)); 
        } 

        line1[i] = Integer.parseInt(digits); 

        i++; 
       } 
      } 

     } 

     int numOfChunks = 10; 

     for (int k = 0; k < numOfChunks; k++) { 
      line2[k] = line1[k]; 
     } 

     for (int r = 0; r < line2.length; ++r) { 
      System.out.print(line2[r] + " "); 
     } 

     System.out.println(); 

     numOfChunks = 9; 
     int j = 10; 

     for (int l = 0; l <= numOfChunks; ++l, ++j) { 
      line3[l] = line1[j]; 
     } 

     for (int r = 0; r < line3.length; ++r) { 
      System.out.print(line3[r] + " "); 
     } 

     System.out.println(); 

     numOfChunks = 6; 
     j = 20; 
     for (int m = 0; m < numOfChunks; ++m, ++j) { 
      line4[m] = line1[j]; 
     } 

     for (int r = 0; r < line4.length; ++r) { 
      System.out.print(line4[r] + " "); 
     } 

     System.out.println(); 

     int line1_maxsum = max_subarray(line2, 1, 10); 
     int line2_maxsum = max_subarray(line3, 11, 20); 
     int line3_maxsum = max_subarray(line4, 21, 26); 

     System.out.println(line1_maxsum); 
     System.out.println(line2_maxsum); 
     System.out.println(line3_maxsum); 

    } 

} 

Answer 1

我不知道是什麼，因爲它的這些亂七八糟的方法max_subarray是幹什麼的，但如果你想在行加起來的數量嘗試在max_subarray方法使用該我猜你可以解決它。

public static int max_subarray(int[] a, int low, int high) { 

    int theMiddle = (int) Math.floor(a.length/2); 
    int leftSum = addArray(a, 0, theMiddle); 
    int rightSum = addArray(a, theMiddle, a.length); 

    if(leftSum > rightSum){ 
     return leftSum; 
    }else{ 
     return rightSum; 
    } 

} 

public static int addArray(int[] a, int start, int end) { 
    int sum = 0; 
    for (int i = start; i < end; i++) { 
     sum = sum + a[i]; 
    } 

    return sum; 
}