Junit测试用例数组矩阵加法和乘法怎么写?
How to write Junit test case array matrix addition and multiplication?
package testMatrix;
public class MatrixAdd {
public int [][] addtionOfArray(int [][] numbers){
int length = numbers.length;
int output[][] = new int[length][length];
for(int i=0; i< length; i++){
for(int j=0; i< length; j++){
output[i][j] = numbers[i][j] + numbers[i][j];
}
}
return output;
}
}
如何编写数组矩阵相加的Junit测试?以及矩阵乘法的 Junit 测试
在 test
源集中你可以使用类似的东西
import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
public class MatrixAddTest {
private MatrixAdd testee = MatrixAdd();
@Test
public void testAdditionOfArray(){
int[][] expected = new int[][]{new int[]{2, 0}, new int[]{0, 2}};
int[][] actual = testee.addtionOfArray(new int[][]{new int[]{1, 0}, new int[]{0, 1}});
Assertions.assertEquals(expected, actual);
}
}
但是,由于 MatrixAdd 看起来像一个实用程序 class,您也可以将这些方法设为静态,您实际上并不需要任何地方的 MatrixAdd 实例。
请查看此示例并阅读评论以更好地理解,我真的希望这段代码对您有用,或者至少让您了解如何
矩阵乘法有效:
/* A naive recursive implementation that simply follows
the above optimal substructure property */
class MatrixChainMultiplication {
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
static int MatrixChainOrder(int p[], int i, int j)
{
if (i == j)
return 0;
int min = Integer.MAX_VALUE;
// place parenthesis at different places between
// first and last matrix, recursively calculate
// count of multiplications for each parenthesis
// placement and return the minimum count
for (int k = i; k < j; k++)
{
int count = MatrixChainOrder(p, i, k)
+ MatrixChainOrder(p, k + 1, j)
+ p[i - 1] * p[k] * p[j];
if (count < min)
min = count;
}
// Return minimum count
return min;
}
// Driver code
public static void main(String args[])
{
int arr[] = new int[] { 1, 2, 3, 4, 3 };
int n = arr.length;
System.out.println(
"Minimum number of multiplications is "
+ MatrixChainOrder(arr, 1, n - 1));
}
}
package testMatrix;
public class MatrixAdd {
public int [][] addtionOfArray(int [][] numbers){
int length = numbers.length;
int output[][] = new int[length][length];
for(int i=0; i< length; i++){
for(int j=0; i< length; j++){
output[i][j] = numbers[i][j] + numbers[i][j];
}
}
return output;
}
}
如何编写数组矩阵相加的Junit测试?以及矩阵乘法的 Junit 测试
在 test
源集中你可以使用类似的东西
import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
public class MatrixAddTest {
private MatrixAdd testee = MatrixAdd();
@Test
public void testAdditionOfArray(){
int[][] expected = new int[][]{new int[]{2, 0}, new int[]{0, 2}};
int[][] actual = testee.addtionOfArray(new int[][]{new int[]{1, 0}, new int[]{0, 1}});
Assertions.assertEquals(expected, actual);
}
}
但是,由于 MatrixAdd 看起来像一个实用程序 class,您也可以将这些方法设为静态,您实际上并不需要任何地方的 MatrixAdd 实例。
请查看此示例并阅读评论以更好地理解,我真的希望这段代码对您有用,或者至少让您了解如何 矩阵乘法有效:
/* A naive recursive implementation that simply follows
the above optimal substructure property */
class MatrixChainMultiplication {
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
static int MatrixChainOrder(int p[], int i, int j)
{
if (i == j)
return 0;
int min = Integer.MAX_VALUE;
// place parenthesis at different places between
// first and last matrix, recursively calculate
// count of multiplications for each parenthesis
// placement and return the minimum count
for (int k = i; k < j; k++)
{
int count = MatrixChainOrder(p, i, k)
+ MatrixChainOrder(p, k + 1, j)
+ p[i - 1] * p[k] * p[j];
if (count < min)
min = count;
}
// Return minimum count
return min;
}
// Driver code
public static void main(String args[])
{
int arr[] = new int[] { 1, 2, 3, 4, 3 };
int n = arr.length;
System.out.println(
"Minimum number of multiplications is "
+ MatrixChainOrder(arr, 1, n - 1));
}
}