This question already has answers here:
What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it?
(26 answers)
Closed 4 years ago.
I'm trying to write a code that multiplies two given matrices together and displays the results. Here is the code I have down so far:
public static void main(String[] args) {
double firstMatrix[][] = {{1, 2}, {3, 4}, {5, 6}};
double secondMatrix[][] = {{1, 2, 3}, {4, 5, 6}};
double[][] thirdMatrix = multiplyMatrix((firstMatrix), (secondMatrix));
System.out.println("The product of the matrices is ");
for (int i = 0; i < thirdMatrix.length; i++) {
for (int j = 0; j < thirdMatrix[0].length; j++) {
System.out.print(firstMatrix[i][j] + " ");
if (i == 1 && j == 2) {
System.out.print(" * ");
} else {
System.out.print(" ");
}
}
for (int j = 0; j < thirdMatrix[0].length; j++) {
System.out.print(secondMatrix[i][j] + " ");
if (i == 1 && j == 2) {
System.out.print(" = ");
} else {
System.out.print(" ");
}
}
for (int j = 0; j < thirdMatrix[0].length; j++) {
System.out.printf("%.1f", thirdMatrix[i][j]);
}
System.out.println();
}
}
public static double[][] multiplyMatrix(double[][] a, double[][] b) {
double c[][] = new double[3][3];
for (int i = 0; i < c.length; i++) {
for (int j = 0; j < b.length; j++) {
for (int k = 0; k < a.length; k++) {
c[i][j] += ((a[i][k]) * (b[k][j]));
}
}
}
return c;
}
But when I try to execute the program I receive an Array Index out of Bounds Error which shouldn't be the case since my for loops are making sure not to step out of bounds from the array. I don't understand where to change my code because it looks fine until I try to execute. Any help is really appreciated.
If 1st matrix's dimensions is mxn and the 2nd's nxp then the product's dimensions mxp. Right?
So this 2nd level of the triple loop should iterate through the 2nd dimension of the product array:
for (int j = 0; j < b.length; j++)
should be:
for (int j = 0; j < b[0].length; j++)
b[0].length is the number of columns of the matrix b.
Also the 3d level of the triple loop should be:
for (int k = 0; k < b.length; k++)
b.length is the number of rows of the matrix b which is equal to the number of columns of the matrix a.
Edit
In a generalized version you could change this:
double c[][] = new double[3][3];
to
double c[][] = new double[a.length][b[0].length];
public static void main(String[] args) {
double firstMatrix[][] = {{1, 2}, {3, 4}, {5, 6}};
double secondMatrix[][] = {{1, 2, 3}, {4, 5, 6}};
double[][] thirdMatrix = multiplyMatrix((firstMatrix), (secondMatrix));
System.out.println("The product of the matrices is ");
for (int i = 0; i < thirdMatrix.length; i++) {
for (int j = 0; j < firstMatrix[0].length; j++) {// check this one
System.out.print(firstMatrix[i][j] + " ");
if (i == 1 && j == 2) {
System.out.print(" * ");
} else {
System.out.print(" ");
}
}
for (int j = 0; j < thirdMatrix.length; j++) {
if (i <= 1) {// you need are condition to stop
System.out.print(secondMatrix[i][j] + " ");
if (i == 1 && j == 2) {
System.out.print(" = ");
} else {
System.out.print(" ");
}
}
}
for (int j = 0; j < thirdMatrix[0].length; j++) {
System.out.print(thirdMatrix[i][j]);
}
System.out.println();
}
}
public static double[][] multiplyMatrix(double[][] a, double[][] b) {
double c[][] = new double[3][3];
for (int i = 0; i < c.length; i++) {
// Change b.length to b[0].length;
for (int j = 0; j < b[0].length; j++) {
// Change a.length to a[0].length;
for (int k = 0; k < a[0].length; k++) {
c[i][j] += ((a[i][k]) * (b[k][j]));
}
}
}
return c;
}
The program should multiply 2 matrices 4x1 and 1x4 and output the result to the console (matrix 4X4). But nothing displays. What's the problem?
public class Matrix {
public static void main(String[] args) {
int[][] matrixA = new int[4][1];
int[][] matrixB = new int[1][4];
int[][] matrixC = new int[4][4];
matrixA[0][0] = 1;
matrixA[1][0] = 2;
matrixA[2][0] = 3;
matrixA[3][0] = 4;
matrixB[0][0] = 4;
matrixB[0][1] = 3;
matrixB[0][2] = 2;
matrixB[0][3] = 1;
for (int i = 0; i < 4; i++) { // A rows
for (int j = 0; j < 4; j++) { // B columns
for (int k = 0; k < 1; k++) { // A columns
matrixC[i][j] += matrixA[i][k] * matrixB[k][j];
System.out.print(matrixC[i][j] + " ");
}
}
}
int j = 0;
for (int i = 0; i < 4; i++) {
for (int k = 0; k < 1; k++)
System.out.print(matrixC[i][j] + " ");
System.out.println();
}
} //end main
} //end class
You introduced a variable j before your second set of for loops. Also, even if they're optional, I would highly recommend always including braces. And k < 4. Like,
for (int i = 0; i < 4; i++) {
for (int k = 0; k < 4; k++) {
System.out.print(matrixC[i][k] + " "); // not [i][j]
}
System.out.println();
}
or just use Arrays.deepToString(Object[]) like
System.out.println(Arrays.deepToString(matrixC));
Hope this helps:
public class Matrix {
public static void main(String[] args) {
int[][] matrixA = {{1}, {2}, {3}, {4}};
int[][] matrixB = {{4, 3, 2, 1}};
int[][] matrixC = new int[4][4];
for (int i = 0; i < 4; i++) { // A rows
for (int j = 0; j < 4; j++) { // B columns
for (int k = 0; k < 1; k++) { // A columns
matrixC[i][j] += matrixA[i][k] * matrixB[k][j];
System.out.print(matrixC[i][j] + " ");
}
}
System.out.println();
}
}
}
Output:
4 3 2 1
8 6 4 2
12 9 6 3
16 12 8 4
I am working on a problem where I've to print the largest sum among all the hourglasses in the array. You can find the details about the problem here-
What I tried:
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for (int arr_i = 0; arr_i < 6; arr_i++) {
for (int arr_j = 0; arr_j < 6; arr_j++) {
arr[arr_i][arr_j] = in.nextInt();
}
}
int sum = 0;
int tmp_sum = 0;
for (int arr_i = 0; arr_i < 4; arr_i++) {
for (int arr_j = 0; arr_j < 4; arr_j++) {
if (arr[arr_i][arr_j] > 0) {
sum = sum + (arr[arr_i][arr_j]) + (arr[arr_i][arr_j + 1]) + (arr[arr_i][arr_j + 2]);
sum = sum + (arr[arr_i + 1][arr_j + 1]);
sum = sum + (arr[arr_i + 2][arr_j]) + (arr[arr_i + 2][arr_j + 1]) + (arr[arr_i + 2][arr_j + 2]);
if (tmp_sum < sum) {
tmp_sum = sum;
}
sum = 0;
}
}
}
System.out.println(tmp_sum);
}
}
Input:
1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 9 2 -4 -4 0
0 0 0 -2 0 0
0 0 -1 -2 -4 0
Output:
12
Expected Output:
13
Screenshot:
I don't know where I'm doing wrong. I cannot understand why the expected output is 13. According to the description given in the problem it should be 10. Is this a wrong question or my understanding about this is wrong?
Remove the if (arr[arr_i][arr_j] > 0) statement. It prevents finding the answer at row 1, column 0, because that cell is 0.
Comments for other improvements to your code:
What if the best hourglass sum is -4? You should initialize tmp_sum to Integer.MIN_VALUE. And name it maxSum, to better describe it's purpose.
You shouldn't define sum outside the loop. Declare it when it is first assigned, then you don't have to reset it to 0 afterwards.
Your iterators should be just i and j. Those are standard names for integer iterators, and keeps code ... cleaner.
If you prefer longer names, use row and col, since that is what they represent.
You don't need parenthesis around the array lookups.
For clarity, I formatted the code below to show the hourglass shape in the array lookups.
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for (int i = 0; i < 6; i++){
for (int j = 0; j < 6; j++){
arr[i][j] = in.nextInt();
}
}
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
int sum = arr[i ][j] + arr[i ][j + 1] + arr[i ][j + 2]
+ arr[i + 1][j + 1]
+ arr[i + 2][j] + arr[i + 2][j + 1] + arr[i + 2][j + 2];
if (maxSum < sum) {
maxSum = sum;
}
}
}
System.out.println(maxSum);
This was my solution. I wrapped an if statement around the code that calculates the sum, that makes sure we don't go out of bounds.
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
int max = Integer.MIN_VALUE;
int tempMax = 0;
for(int i=0; i < 6; i++){
for(int j=0; j < 6; j++){
arr[i][j] = in.nextInt();
}
}
for(int i=0; i < 6; i++){
for(int j=0; j < 6; j++){
if (i + 2 < 6 && j + 2 < 6) {
tempMax += arr[i][j] + arr[i][j + 1] + arr[i][j + 2];
tempMax += arr[i + 1][j + 1];
tempMax += arr[i + 2][j] + arr[i + 2][j + 1] + arr[i + 2][j + 2];
if (max < tempMax) {
max = tempMax;
}
tempMax = 0;
}
}
}
System.out.println(max);
}
Here's the simple and easy to understand C# equivalent code for your hourglass problem.
class Class1
{
static int[][] CreateHourGlassForIndex(int p, int q, int[][] arr)
{
int[][] hourGlass = new int[3][];
int x = 0, y = 0;
for (int i = p; i <= p + 2; i++)
{
hourGlass[x] = new int[3];
int[] temp = new int[3];
int k = 0;
for (int j = q; j <= q + 2; j++)
{
temp[k] = arr[i][j];
k++;
}
hourGlass[x] = temp;
x++;
}
return hourGlass;
}
static int findSumOfEachHourGlass(int[][] arr)
{
int sum = 0;
for (int i = 0; i < arr.Length; i++)
{
for (int j = 0; j < arr.Length; j++)
{
if (!((i == 1 && j == 0) || (i == 1 && j == 2)))
sum += arr[i][j];
}
}
return sum;
}
static void Main(string[] args)
{
int[][] arr = new int[6][];
for (int arr_i = 0; arr_i < 6; arr_i++)
{
string[] arr_temp = Console.ReadLine().Split(' ');
arr[arr_i] = Array.ConvertAll(arr_temp, Int32.Parse);
}
int[] sum = new int[16];
int k = 0;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++)
{
int[][] hourGlass = CreateHourGlassForIndex(i, j, arr);
sum[k] = findSumOfEachHourGlass(hourGlass);
k++;
}
}
//max in sum array
Console.WriteLine(sum.Max());
}
}
Happy Coding.
Thanks,
Ankit Bajpai
You can try this code:
I think this will be easy to understand for beginners.
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for(int arr_i=0; arr_i < 6; arr_i++){
for(int arr_j=0; arr_j < 6; arr_j++){
arr[arr_i][arr_j] = in.nextInt();
}
}
int sum = 0;
int sum2 = 0;
int sum3 = 0;
int x = 0;
int max = Integer.MIN_VALUE;
for(int i = 0; i < 4; i++){
for(int j = 0; j < 4; j++){
for(int k = 0; k < 3; k++){
sum += arr[i][j+k]; //top elements of hour glass
sum2 += arr[i+2][j+k]; //bottom elements of hour glass
sum3 = arr[i+1][j+1]; //middle elements of hour glass
x = sum + sum2 + sum3; //add all elements of hour glass
}
if(max < x){
max = x;
}
sum = 0;
sum2 = 0;
sum3 = 0;
x = 0;
}
}
System.out.println(max);
}
}
Here is another easy option, hope it helps:
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 a[][] = new int[6][6];
for(int i=0; i < 6; i++){
for(int j=0; j < 6; j++){
a[i][j] = in.nextInt();
}
}
int hg = Integer.MIN_VALUE, sum;
for(int i=0; i<4; i++){
for(int j=0; j<4; j++){
sum = 0;
sum = sum + a[i][j] + a[i][j+1] + a[i][j+2];
sum = sum + a[i+1][j+1];
sum = sum + a[i+2][j] + a[i+2][j+1] + a[i+2][j+2];
if(sum>hg)
hg = sum;
}
}
System.out.println(hg);
in.close();
}
}
there is another opetion in case of -(minus) and zero output we can use shorted ser Treeset for the same . below is the sameple code
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for(int i=0; i < 6; i++){
for(int j=0; j < 6; j++){
arr[i][j] = in.nextInt();
}
}
int sum=0;int output=0;
Set<Integer> set=new TreeSet<Integer>();
for(int k=0;k<4;k++ )
{
for(int y=0;y<4;y++)
{
sum=arr[k][y]+arr[k][y+1]+arr[k][y+2]+arr[k+1][y+1]+arr[k+2][y]+arr[k+2][y+1]+arr[k+2][y+2]; set.add(sum);
}
}
int p=0;
for(int u:set)
{
p++;
if(p==set.size())
output=u;
}
System.out.println(output);
}
}
Solved in PHP, may be helpful.
<?php
$handle = fopen ("php://stdin","r");
$input = [];
while(!feof($handle))
{
$temp = fgets($handle);
$input[] = explode(" ",$temp);
}
$maxSum = PHP_INT_MIN;
for($i=0; $i<4; $i++)
{
for($j=0; $j<4; $j++)
{
$sum = $input[$i][$j] + $input[$i][$j + 1] + $input[$i][$j + 2]
+ $input[$i + 1][$j + 1] +
$input[$i + 2][$j] + $input[$i + 2][$j + 1] + $input[$i + 2][$j + 2];
if($sum > $maxSum)
{
$maxSum = $sum;
}
}
}
echo $maxSum;
?>
Passes all test cases
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner read = new Scanner(System.in);
int rowSize = 6;
int colSize = 6;
int[][] array = new int[rowSize][colSize];
for(int row = 0; row < rowSize; row++) {
for(int col = 0; col < colSize; col++) {
array[row][col] = read.nextInt();
}
}
read.close();
int max = Integer.MIN_VALUE;
for(int row = 0; row < 4; row++) {
for(int col = 0; col < 4; col++) {
int sum = calculateHourglassSum(array, row, col);
if(sum > max) {
max = sum;
}
}
}
System.out.println(max);
}
private static int calculateHourglassSum(int[][] array, int rowIndex, int colIndex) {
int sum = 0;
for(int row = rowIndex; row < rowIndex + 3; row++) {
for(int col = colIndex; col < colIndex + 3; col++) {
if(row == rowIndex + 1 && col != colIndex + 1) {
continue;
}
sum += array[row][col];
}
}
return sum;
}
}
function galssSum(array) {
let maxGlass = 0;
if (array[0].length == 3) {
maxGlass = 1;
} else if (array[0].length > 3) {
maxGlass = array.length - 2;
}
let maxValue = -100000;
for (let i = 0; i < maxGlass; i++) {
for (let j = 0; j < maxGlass; j++) {
let a = array[i][j] + array[i][j + 1] + array[i][j + 2];
let b = array[i + 1][j + 1];
let c = array[i + 2][j] + array[i + 2][j + 1] + array[i + 2][j + 2];
let sum = a + b + c;
if (maxValue<sum) {
maxValue = sum;
}
}
}
return maxValue;
}
console.log(galssSum([[1, 1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], [1, 1, 1, 0, 0, 0], [0, 0, 2, 4, 4, 0], [0, 0, 0, 2, 0, 0], [0, 0, 1, 2, 4, 0]]));
int hourglassSum(vector<vector<int>> vec) {
int res = 0;
int size = ((vec[0].size())-2) * ((vec.size())-2);
//cout<<size<<endl;
vector<int> res_vec(size);
int j = 0;
int itr =0 ;
int cnt = 0;
int mid = 0;
int l =0;
while((l+2) < vec.size())
{
while((j+2) < vec.size())
{
for(int i =j ;i<j+3; i+=2)
{
//cout<<i<<" :";
for(int k=l;k<l+3;k++)
{
//cout<<k<<" ";
res_vec[itr] += vec[i][k];
}
//cout<<endl;
}
res_vec[itr] += vec[j+1][l+1];
//cout<<endl;
itr++;
j++;
}
l++;
j=0;
}
int max=res_vec[0];
for(int i =1;i<res_vec.size();i++)
{
if(max < res_vec[i])
{
max = res_vec[i];
}
//cout<<res_vec[i]<< " ";
}
res = max;
//cout<<endl;
return res;
}
// Complete the hourglassSum function below.
static int hourglassSum(int[][] arr) {
int max = Integer.MIN_VALUE;
for (int i = 0; i < arr.length - 2; i++) {
for (int j = 0; j < arr.length - 2; j++) {
int hourGlassSum = (arr[i][j] + arr[i][j + 1] + arr[i][j + 2])
+ (arr[i + 1][j + 1])
+ (arr[i + 2][j] + arr[i + 2][j + 1] + arr[i + 2][j + 2]);
max = Math.max(hourGlassSum,max);
}
}
return max;
}
public static int hourglassSum(List<List<Integer>> arr) {
// Write your code here
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
int sum = arr.get(i).get(j) +arr.get(i).get(j+1) +
arr.get(i).get(j+2)+arr.get(i+1).get(j+1)+
arr.get(i+2).get(j)+arr.get(i+2).get(j+1)+arr.get(i+2).get(j+2);
if (maxSum < sum) {
maxSum = sum;
}
}
}
return maxSum;
}
}
Iterative way,Passing all test cases in hackerank web
public static int hourglassSum(List<List<Integer>> arr) {
// Write your code here
int rowsCount=arr.size();
int colCount=arr.get(0).size();
Integer max=Integer.MIN_VALUE;
Integer subSum=0;
for(int r=0; (r+3)<=rowsCount; r++)
{
for(int c=0; (c+3)<=colCount; c++)
{
subSum= hourglassSubSum(arr,r,c);
System.out.println("r,c,subSum "+r+" "+c+" "+" "+subSum);
if(subSum>max)
{
max=subSum;
}
}
}
return max;
}
public static int hourglassSubSum(List<List<Integer>> hourglassArray,
int rowIndex,int colIndex) {
// Write your code here
Integer subSum=0;
for(int i=rowIndex;i<(rowIndex+3);i++)
{
for(int j=colIndex;j<(colIndex+3);j++)
{
if(i==(rowIndex+1) && (j==colIndex || j==colIndex+2))
{
continue;
}
subSum=subSum+hourglassArray.get(i).get(j);
}
}
return subSum;
}
Solution for actual "2D Array - DS" challenge from HackerRank https://www.hackerrank.com/challenges/2d-array
public static int hourglassSum(List<List<Integer>> arr) {
int maxSum = Integer.MIN_VALUE;
for (int col=0; col <= 3; col++) {
for (int row=0; row <= 3; row++) {
int sum = calcHourglass(arr, col, row);
maxSum = Math.max(sum, maxSum);
}
}
return maxSum;
}
private static int calcHourglass(List<List<Integer>> arr, int col, int row) {
int sum = 0;
for (int i=0; i < 3; i++) {
sum += arr.get(row).get(col+i); // the top of the hourglass
sum += arr.get(row+2).get(col+i); // the bottom of the hourglass
}
sum += arr.get(row+1).get(col+1); // the center
return sum;
}
import java.io.*;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int low = -9,high = 5;
int lh = low * high;
int sum = 0, i, j;
int max = 0;
int a[][] = new int[6][6];
for (i = 0; i < 6; i++) {
for (j = 0; j < 6; j++) {
a[i][j] = in.nextInt();
}
}
for (i = 0; i < 4; i++) {
for (j = 0; j < 4; j++) {
sum = (a[i][j] + a[i][j+1] + a[i][j+2]);
sum = sum + a[i+1][j+1];
sum = sum + (a[i+2][j] + a[i+2][j+1] + a[i+2][j+2]);
if (sum > lh) lh = sum;
}
}
System.out.print(lh);
}
}
Here you go..
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a[][] = new int[6][6];
int max = 0;
for (int i = 0; i < 6; i++) {
for (int j = 0; j < 6; j++) {
a[i][j] = in.nextInt();
}
}
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
int sum = a[i][j] + a[i][j + 1] + a[i][j + 2] + a[i + 1][j + 1]
+ a[i + 2][j] + a[i + 2][j + 1] + a[i + 2][j + 2];
if (sum > max || (i == 0 && j == 0)) {
max = sum;
}
}
}
System.out.println(max);
}
I try to convert 1D to 2D array, but I keep getting java.lang.ArrayIndexOutOfBoundsException, and I have tried whatever I could find on the stackoverflow or internet, but I do not understand why I have this issue?
public class Arrayto2DArray {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
int[] a = {0,1, 6, 83, 4, 5, 12, 7};
int[][] b = new int[4][4];
for (int i = 0; i < b.length; i++) {
for (int j = 0; j < b[i].length; j++) {
b[i][j]=0;
System.out.print(b[i][j]);
}
System.out.println();
}
System.out.println("--------------------------");
for (int i = 0; i < 4; i++) {
for (int j = 0; j < b[i].length; j++) {
try{
b[i][j] = a[i+j*4];
}catch(Exception e){
e.printStackTrace();
System.out.println(e);
}
}
}
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
System.out.print(b[i][j]);
}
System.out.println();
}
}
}
I kind of know why I get this error and it because of this line
b[i][j] = a[i+j*4];
but I cannot come up any formula better than this.
Consider the second for-loop
Lets say when i = 3 and j= 3
a[i+j*4] evaluates to a[15] which is out of the array
When you declare you 2-d array, you specified int[][] b = new int[4][];, meaning that the first inner for loop for (int j = 0; j < b[i].length; j++) should result in a NullPointerException since b[i].length has no predefined length. Before intering the inner for loop, you should define the size of each b[i] like b[i] = new int[somenumber]
In regards to convert the 1d loop to a 2d, you need to define the rule around spliting it into the 2-d array. Then accordingly the second for loop need modification
EDIT:You modified your code to have an int[4][4] array, which means you have 16 placeholders. Your 1-d array contain only 8 placeholders. It depends on how you want to sort the array, like it can be
b 0 1 2 3
0 0 1 6 83
1 4 5 12 7
2 0 0 0 0
3 0 0 0 0
or any other pattern
Assuming the length of 1-d array is 8 and the total index of 2-d array is 16, the following is more of a general solution:
public static void main(String[] args) {
// TODO code application logic here
int[] a = { 0, 1, 6, 83, 4, 5, 12, 7 };
int[][] b = new int[4][4];
for (int i = 0; i < b.length; i++) {
for (int j = 0; j < b[i].length; j++) {
b[i][j] = 0;
System.out.print(b[i][j] + "\t");
}
System.out.println();
}
System.out.println("--------------------------");
for (int i = 0; i < 4; i++) {
for (int j = 0; j < b[i].length; j++) {
try {
if ((j + i * 4) < a.length)
b[i][j] = a[j + i * 4];
} catch (Exception e) {
e.printStackTrace();
System.out.println(e);
}
}
}
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
System.out.print(b[i][j] + "\t");
}
System.out.println();
}
}
What is wrong is your 2D array is a 4*4, so that means for each index, there will be 4 elements i.e. say values. So there would be a total of 16 values, so you would need a 1D array with 16 elements for the above code to work. Right now you have only 8 values in your 1D array.
I think your code should change like this
public class Arrayto2DArray {
public static void main(String[] args) {
// TODO code application logic here
int[] a = {0,1, 6, 83, 4, 5, 12, 7};
int[][] b = new int[4][2];
for (int i = 0; i < b.length; i++) {
for (int j = 0; j < b[i].length; j++) {
b[i][j]=0;
System.out.print(b[i][j]);
}
System.out.println();
}
System.out.println("--------------------------");
for (int i = 0; i < 4; i++) {
for (int j = 0; j < b[i].length; j++) {
try{
b[i][j] = a[i+j*4];
}catch(Exception e){
e.printStackTrace();
System.out.println(e);
}
}
}
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 2; j++) {
System.out.print(b[i][j]+" ");
}
System.out.println();
}
}
}
Your correct formula should be like.
int[] a = {0,1, 6, 83, 4, 5, 12, 7};
int[][] b = new int[4][2];
int k=0;
for (int i = 0; i < b.length; i++) { // Column of 2D array
for (int j = 0; j < b[0].length; j++) { // Row of 2D array
b[i][j]= a[k++];
System.out.println(b[i][j]);
}
}
Edit:
For general case, If your row is fixed at 4 but column is not fixed than your 2D conversion formula should be like below.
int col = a.length / 4;
int remainder = a.length % 4;
if (remainder > 0) {
col = col + 1; // Get the correct column size.
}
int[][] b = new int[4][col];
int k = 0;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < col; j++) {
if (k < a.length) {
b[i][j] = a[k];
System.out.println(b[i][j]);
}
k++;
}
}
I tried to run this code as a score table where I have chars and ints as below for heading;
A B c
1
2 65 //this is where I'm stuck again!
3
In order to print the score like (65) above in a particular place (matrix) but as soon as I try to add the print statements the table falls apart. Any help would be appreciated;
public class Table3 {
static int[][] list = new int[4][4];
//private char column = 'A';
//private int row = 1;
private static int row = 1;
public Table3(){
//column = 'A';
for (int i = 0; i < 4; i++) {
for (int j = 1; j < 4; j++)
list[i][j] = 0;
}
}
public static void table(char col, int row, int value) {
//System.out.printf("\n\n%s\n", "Table");
for (int i = 1; i < 4; i++) {
System.out.print(row + " ");
row++;
for (int j = 1; j < 4; j++)System.out.print(col + " ");
System.out.println("\n");
col++;
if (row >= 0 && row <= 4 && col >=0 && col <= 4)
System.out.print(list[col][row]=value);
System.out.println("\n");
}
}
}
Client
public class TableTest {
public static void main(String[] args) {
// TODO Auto-generated method stub
Table3 t = new Table3();
t.table('A', 5, 5);
}
}
Learn how to use System.out.printf. The docs are here.
public class Table3
{
static int numRows = 4;
static int numCols = 4;
static int[][] list = new int[numRows][numCols];
public Table3()
{
//column = 'A';
for (int i = 0; i < 4; i++)
{
//this is the row number so you don't have to print it manually
//just print the array
list[i][0] = i;
//initialize the list to 0
for (int j = 1; j < 4; j++)
{
list[i][j] = 0;
}
}
}
public static void table(char col, int row, int value)
{
list[row][col] = value;
int columnWidth = 5; //in characters
//empty space before first column header
for (int i = 0; i < columnWidth; i++)
{
System.out.print(" ");
}
//print the column headers (A through C)
for (int i = 1; i < numCols; i++)
{
System.out.printf("%-" + columnWidth" + "c", (char)(64 + i));
}
System.out.println(); //get off of the column header row
//print the rest of the table
for (int i = 1; i < numRows; i++)
{
for (int j = 0; j < numCols; j++)
{
if (list[i][j] == 0)
{
System.out.printf("%" + columnWidth + "s", " ");
}
else
{
System.out.printf("%-" + columnWidth + "d", list[i][j]);
}
}
System.out.println("\n");
}
}
}