在數組中查找差異

Question

我不確定如何設置要存儲在數組差異中的差異。存儲的數字應該是5-（1 + 2 + 3），7-（1,2,4），8-（3,5,9）：輸出應該是差值[0] = 1，差值[1] = 0，差異[2] = 9

import java.util.Scanner; 

    public class Main { 
     public static int[][] Array = { { 5, 1, 2, 3 }, { 7, 1, 2, 4 }, { 8,3,5,9 } }; //My 2D array// 
     int [] differences = new int [3]; 

     public static int[] Sum(int[][] array) { 
     int index = 0; //setting the index to 0// 
     int temp[] = new int[array[index].length]; //making a temperary variable// 
     for (int i = 0; i < array.length; i++) { 
      int sum = 0; 
      for (int j = 1; j < array[i].length; j++) { 
      sum += array[i][j]; //going to add the rows after the first column// 
      } 
      temp[index] = sum; 

      for(int a = 0; a<differences.length; a++){ 
      if(sum != array[index][0]) 
       sum -= array[i][j]; 




       System.out.println("the first integer " + array[index][0] + " the answer is " + sum); //going to print out the first integer each row and print out the sum of each row after the first column// 
      index++; //index is going to increment// 
      } 
      return temp; 
     } 

     public static void main(String[] args) { 
      new Main().Sum(Array); 
     } 
     } 

輸出：

the first integer 5 the answer is 6 
    the first integer 7 the answer is 7 
    the first integer 8 the answer is 17 

Answer 1

你爲什麼要當它是這種簡單的你的任務複雜化？ :)

public int[] Sum(int[][] array) 
    { 
    int sum; 
    for(int i = 0; i < Array.length; i++) 
     { 
     sum = Array[i][0] * -1; 

     for(int j = 1; j < Array[i].length; j++) 
      { 
      sum += Array[i][j]; 
      } 

     differences[i] = sum; 
     } 

    return differences; 
    } 

Answer 2

如果我正確理解你的問題，我認爲，你想要把一個

differences[i] = Array[i][0] - sum 

地方在你的代碼