我不确定如何设置差异以存储在数组差异中。存储的数字应该是 5-(1+2+3), 7-(1,2,4), 8-(3,5,9) :输出应该是differences[0]= 1,differences[1] = 0,差异[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);
}
}
输出:
the first integer 5 the answer is 6
the first integer 7 the answer is 7
the first integer 8 the answer is 17
最佳答案
既然任务这么简单,你为什么要把它复杂化呢? :)
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;
}
关于java - 查找数组中的差异,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21582142/