Java 在二维数组中添加对角线值
我有以下2d数组Java 在二维数组中添加对角线值,java,arrays,2d,Java,Arrays,2d,我有以下2d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
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)
公共静态int-arraySum(int[][]数组)
{
int-total=0;
for(int row=0;row
公共静态int-arraySum(int[][]数组)
{
int-total=0;
for(int index=0;index
当然,这假设尺寸为m x m。因为对角线是完全正方形的,所以只需要一个循环就可以添加对角线
从orgin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;
}
公共静态int-arraySum(int[][]数组){
int-total=0;
for(int row=0;row
添加两条对角线: 从原点添加对角线:(注意,它将中心添加两次..如果需要,可以减去一)
公共静态int-arraySum(int[][]数组){
int-total=0;
for(int row=0;row
导入java.util.Scanner;
公共类解决方案{
公共静态void main(字符串[]args){
扫描仪输入=新扫描仪(系统输入);
int n=in.nextInt();
int leftStartDiagnol=0;
int rightStartDiagnol=n;
int leftTotal=0;
int rightotal=0;
int a[][]=新int[n][n];
for(int a_i=0;a_i
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);
}
}
import java.io.*;
导入java.util.*;
导入java.text.*;
导入java.math.*;
导入java.util.regex.*;
公共类解决方案{
公共静态void main(字符串[]args){
扫描仪输入=新扫描仪(系统输入);
int n=in.nextInt();
int a[][]=新int[n][n];
for(int a_i=0;a_iimportjava.io.*;
导入java.util.*;
公共级数字图书馆{
公共静态void main(字符串[]args){
扫描仪s=新的扫描仪(System.in);
int n=Integer.parseInt(s.nextLine());//数组大小
int[][]a=新int[n][n];//新数组
对于(inti=0;i这是我的解决方案。请查看它
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));
}
import java.io.*;
导入java.util.*;
导入java.text.*;
导入java.math.*;
导入java.util.regex.*;
公共类解决方案{
公共静态void main(字符串[]args){
扫描仪输入=新扫描仪(系统输入);
int n=in.nextInt();
int a[][]=新int[n][n];
整数total1=0;
整数2=0;
for(int a_i=0;a_i
}PHP解决方案。逻辑与此处已发布的完全相同。只有在PHP中不同
<?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";
?>
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);
int-total=0;
for(int row=0;row
这是我用来求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;
}
static int diagSum(int[]arr){
int-pd=0;
int-sd=0;
整数和=0;
对于(int i=0;i如果您想在二维数组中添加两个对角线,那么下面是我的程序
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;
}
静态int对角线差(int[]arr){
int r=阵列长度;
int sum1=0;
int-sum2=0;
int j=0;
对于(int i=0;i其对角线之和在kotlin中的差值:
val total = (0 until array.size).sumBy { array[it][it] - array[it][array.size - it - 1] }
如果您想要绝对差异:
return Math.abs(total)
公共类矩阵{
公共静态void main(字符串[]args){
整数和=0;
int a[][]=新int[][{{2,3,4},{4,5,6},{5,5,5};
对于(int i=0;i如果要在二维数组中添加两个对角线,则有一个解决方案:
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--;
}
int j=row\u col\u size-1;
for(int i=0;i
虽然这篇文章很老。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
}
func对角线差分(arr[]]int32)int32{
返回int32(math.Abs(float64(arraySumPrimary(arr))-float64(arraySumSecondary(arr)))
}
func数组通用(arr[][]int32)int32{
变量总计int32=0
对于行:=0;行=0&&y
D
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;
}
val total = (0 until array.size).sumBy { array[it][it] - array[it][array.size - it - 1] }
return Math.abs(total)
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);
}
}
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--;
}
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
}