在二维数组中添加对角线值

Question

我有以下二维数组

        int [][] array = {{ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9},
                          {10, 11, 12, 13, 14, 15, 16, 17, 18, 19},
                          {20, 21, 22, 23, 24, 25, 26, 27, 28, 29},
                          {30, 31, 32, 33, 34, 35, 36, 37, 38, 39},
                          {40, 41, 42, 43, 44, 45, 46, 47, 48, 49},
                          {50, 51, 52, 53, 54, 55, 56, 57, 58, 59},
                          {60, 61, 62, 63, 64, 65, 66, 67, 68, 69},
                          {70, 71, 72, 73, 74, 75, 76, 77, 78, 79},
                          {80, 81, 82, 83, 84, 85, 86, 87, 88, 89},
                          {90, 91, 92, 93, 94, 95, 96, 97, 98, 99}};

我有这段代码来查找数组中所有值的总和。 如何修改它以仅添加从 0 开始的对角线值（0+11+22+33 等）？

 public static int arraySum(int[][] array)
{
    int total = 0;

    for (int row = 0; row < array.length; row++)
    {
        for (int col = 0; col < array[row].length; col++)
            total += array[row][col];
    }

    return total;
}

Answer 1

由于对角线是完美的正方形，因此您只需要一个循环来添加对角线。

---

从原点添加对角线：

public static int arraySum(int[][] array){
    int total = 0;

    for (int row = 0; row < array.length; row++)
    {

        total += array[row][row];
    }

    return total;
}

---

添加两条对角线：

从原点添加对角线：（注意它添加了两次中心......如果需要，您可以减去一个）

public static int arraySum(int[][] array){
    int total = 0;

    for (int row = 0; row < array.length; row++)
    {
        total += array[row][row] + array[row][array.length - row-1];
    }

    return total;
}

Answer 2

public static int arraySum(int[][] array)
{
    int total = 0;

    for (int index = 0; index < array.length; index++)
    {
            total += array[index][index];
    }

    return total;
}

这当然假设尺寸为 mxm。

Answer 3

import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int leftStartDiagnol = 0;
        int rightStartDiagnol = n;
        int leftTotal = 0;
        int rightTotal = 0;
        int a[][] = new int[n][n];
        for (int a_i = 0; a_i < n; a_i++) {
            for (int a_j = 0; a_j < n; a_j++) {
                a[a_i][a_j] = in.nextInt();
            }
        }
        for (int a_i = 0; a_i < n; a_i++) {
            boolean leftNotFound = true;
            boolean rightNotFound = true;
            rightStartDiagnol = --rightStartDiagnol;

            for (int a_j = 0; a_j < n; a_j++) {
                if (leftStartDiagnol == a_j && leftNotFound) {
                    leftTotal = leftTotal + a[a_i][a_j];
                    leftNotFound = false;
                }
                if (rightStartDiagnol == a_j && rightNotFound) {
                    rightTotal = rightTotal + a[a_i][a_j];
                    rightNotFound = false;
                }
            }
            leftStartDiagnol = ++leftStartDiagnol;
        }
        int data = leftTotal - rightTotal;
        System.out.println(Math.abs(data));
    }
}

Answer 4

这是我的解决方案 -

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int n = in.nextInt();
    int a[][] = new int[n][n];
    for(int a_i=0; a_i < n; a_i++){
        for(int a_j=0; a_j < n; a_j++){
            a[a_i][a_j] = in.nextInt();
        }
    }
    int l_sum = 0;
    for(int i = 0; i<n ; i++){
        l_sum+=a[i][i];
    }
    int r_sum = 0;
    for(int j = 0; j<n ; j++){
        r_sum+=a[j][n-1-j];
    }

    int sum = l_sum + r_sum;

    System.out.println(sum);
   }
 }

Answer 5

import java.io.*;
import java.util.*;
public class DigArraySum {
public static void main(String[] args) {
    Scanner s=new Scanner(System.in);
    int n=Integer.parseInt(s.nextLine()); //array size
    int[][] a=new int[n][n];  // new array
    for (int i=0;i<n;i++) {
      String str=s.nextLine();
      String[] tempArray=str.split(" ");
      for (int j=0;j<n;j++) {
        a[i][j]=Integer.parseInt(tempArray[j]);
      }
    }
    int x=0;//primary diagonal sum
    int y=0;//secondary diagonal sum
    int sum=0;//total sum
    for (int i=0;i<n;i++) {
      x += a[i][i];
    }
    for (int p=0;p<n;p++) {
      int k=a.length-p-1;
      y+=a[p][a.length-p-1];
      k+=-1;
    }
    sum=x-y;
    System.out.println(Math.abs(sum));
    }
  }

说明：例如3*3矩阵：3//数组大小

11 2 4

4 5 6

10 8 -12

主对角线为：11 5 -12

主对角线总和：11 + 5 - 12 = 4

次对角线是：4 5 10次对角线的总和：4 + 5 + 10 = 19

总和：|4 + 19| = 23

Answer 6

这是我的解决方案。 看看吧。

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int n = in.nextInt();
    int a[][] = new int[n][n];
    int total1=0;
    int total2=0;

    for(int a_i=0; a_i < n; a_i++){
        for(int a_j=0; a_j < n; a_j++){
            a[a_i][a_j] = in.nextInt();
        }
        total1+= a[a_i][a_i];
        total2+=a[a_i][n-1-a_i];
    }
    System.out.println(Math.abs(total1-total2));
}

}

Answer 7

PHP 中的解决方案。 逻辑与此处已发布的逻辑完全相同。 仅在 PHP 中有所不同。

<?php

$handle = fopen ("php://stdin", "r");
function diagonalDifference($a) {
    $sumA = [];
    $sumB = [];

    $n = count($a);

    for ($i = 0; $i < $n; $i++) {
        for ($j = 0; $j < $n; $j++) {
            if ($i === $j) {
                $sumA[] = $a[$i][$j];
            }

            if ($i + $j == count($a) -1 ) {
                $sumB[] = $a[$i][$j];
            }
        }
    }

    $sum1 = array_sum($sumA);
    $sum2 = array_sum($sumB);

    return abs($sum1 - $sum2);
}

fscanf($handle, "%i",$n);
$a = array();
for($a_i = 0; $a_i < $n; $a_i++) {
    $a_temp = fgets($handle);
    $a[] = explode(" ",$a_temp);
    $a[$a_i] = array_map('intval', $a[$a_i]);
}
$result = diagonalDifference($a);
echo $result . "\n";
?>

Answer 8

解决方案是：

int total = 0;
for (int row = 0; row < array.length; row++)
{
    total += array[row][row] + array[row][array.length - row - 1];
}

System.out.println("FINAL ANSWER: " + total);

Answer 9

这是我用来查找主对角线数和次对角线数之和的代码

static int diagSum(int[][] arr) {
    int pd=0;
    int sd=0;
    int sum=0; 
    for(int i=0;i<=arr.length-1;i++){
       pd=pd+arr[i][i];
    } 

    for (int k=0,l=arr.length-1; k<arr.length&&l>=0 ; k++,l--) {
        sd=sd+arr[k][l];
    }

    sum=pd+sd;
    return sum;
}

Answer 10

如果你想在二维数组中添加两个对角线，那么这是我的程序。

static int diagonalDifference(int[][] arr) {


    int r = arr.length;     
    int sum1 = 0;
    int sum2 = 0;

    int j=0;
    for(int i=0;i<r;i++){
        sum1 = sum1 + arr[i][j];
        j++;
    }
    j--;
    for(int i=0;i<r;i++) {
        sum2 = sum2 + arr[i][j];
        j--;
    }
    int sum=sum1+sum2; 
    return sum;
}

Answer 11

kotlin 中其对角线之和之间的差异：

val total = (0 until array.size).sumBy { array[it][it] - array[it][array.size - it - 1] }

如果你想要绝对差异：

return Math.abs(total)

Answer 12

public class matrix {

    public static void main(String[] args) {
        int sum=0;
        int a[][]= new int[][]{{2,3,4},{4,5,6},{5,5,5}};

        for(int i=0;i<=2;i++)
        {
            sum=sum+a[i][i];
        }

        System.out.println(sum);
    }

}

Answer 13

如果你想在二维数组中添加两个对角线，那么这里是一个解决方案：

    int j = row_col_size-1;
    for (int i = 0; i < intArray.length; i++) {
        sum_left += intArray[i][i];
        sum_right += intArray[i][j];
        j--;
    }

Answer 14

虽然这篇文章很老了。 提供我的解决方案。 这可能对某人有用

func diagonalDifference(arr [][]int32) int32 {


return int32(math.Abs(float64(arraySumPrimary(arr)) - float64(arraySumSecondary(arr))) )

}
func arraySumPrimary(arr [][]int32) int32{
    var total int32 =0
    for row :=0; row < len(arr) ; row ++{
        total = total + arr[row][row]
    }
    return total
}

func arraySumSecondary(arr [][]int32) int32{
    var total int32 =0
    x := len(arr)-1 
    y := 0
    for x >=0 && y < len(arr){
        total = total + arr[x][y]
        x--
        y++
    }
    return total
}