重构 java 代码的问题
Problems with refactoring java code
我现在正在做作业,有一个关于在 Java 中重构我的代码的问题。
我现在正在玩数独,我需要检查 3x3 框是否有效。为此,我创建了一个包含所有框数的一维数组,然后比较它们的值。它现在正在工作,但实际上根本没有重构。我真的很想知道是否有任何方法可以减少所有这些复制粘贴。
public static boolean validFieldParts() {
int counter = 0;
boolean isValid = false;
int[] copyArray1 = new int[field.length];
int[] copyArray2 = new int[field.length];
int[] copyArray3 = new int[field.length];
int[] copyArray4 = new int[field.length];
int[] copyArray5 = new int[field.length];
int[] copyArray6 = new int[field.length];
int[] copyArray7 = new int[field.length];
int[] copyArray8 = new int[field.length];
int[] copyArray9 = new int[field.length];
// copy the array
// 1 große Feld
for (int i = 0; i < field.length / 3; i++) {
for (int j = 0; j < field[i].length / 3; j++) {
copyArray1[i * 3 + j] = field[i][j];
}
}
// 2 große Feld
for (int i = 0; i < field.length / 3; i++) {
for (int j = 3; j < 6; j++) {
copyArray2[i * 3 + j - 3] = field[i][j];
}
}
// 3 große Feld
for (int i = 0; i < field.length / 3; i++) {
for (int j = 6; j < 9; j++) {
copyArray3[i * 3 + j - 6] = field[i][j];
}
}
// 4 große Feld
for (int i = 3; i < 6; i++) {
for (int j = 0; j < field[i].length / 3; j++) {
copyArray4[(i - 3) * 3 + j] = field[i][j];
}
}
// 5 große Feld
for (int i = 3; i < 6; i++) {
for (int j = 3; j < 6; j++) {
copyArray5[(i - 3) * 3 + j - 3] = field[i][j];
}
}
// 6 große Feld
for (int i = 3; i < 6; i++) {
for (int j = 6; j < 9; j++) {
copyArray6[(i - 3) * 3 + j - 6] = field[i][j];
}
}
// 7 große Feld
for (int i = 6; i < 9; i++) {
for (int j = 0; j < field[i].length / 3; j++) {
copyArray7[(i - 6) * 3 + j] = field[i][j];
}
}
// 8 große Feld
for (int i = 6; i < 9; i++) {
for (int j = 3; j < 6; j++) {
copyArray8[(i - 6) * 3 + j - 3] = field[i][j];
}
}
// 9 große Feld
for (int i = 6; i < 9; i++) {
for (int j = 6; j < 9; j++) {
copyArray9[(i - 6) * 3 + j - 6] = field[i][j];
}
}
Arrays.sort(copyArray1);
Arrays.sort(copyArray2);
Arrays.sort(copyArray3);
Arrays.sort(copyArray4);
Arrays.sort(copyArray5);
Arrays.sort(copyArray6);
Arrays.sort(copyArray7);
Arrays.sort(copyArray8);
Arrays.sort(copyArray9);
for (int i = 1; i < copyArray1.length; i++) {
if (copyArray1[i] == copyArray1[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray2.length; i++) {
if (copyArray2[i] == copyArray2[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray3.length; i++) {
if (copyArray3[i] == copyArray3[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray4.length; i++) {
if (copyArray4[i] == copyArray4[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray5.length; i++) {
if (copyArray5[i] == copyArray5[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray6.length; i++) {
if (copyArray6[i] == copyArray6[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray7.length; i++) {
if (copyArray7[i] == copyArray7[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray8.length; i++) {
if (copyArray8[i] == copyArray8[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray9.length; i++) {
if (copyArray9[i] == copyArray9[i - 1])
counter++;
else
continue;
}
if (counter > 0)
isValid = false;
else
isValid = true;
return isValid;
}
这里是一个完整的重构。这里的改进:
- 创建了两个新方法:
createCopy
和 isValid
- 删除了未使用的变量
counter
和 isValid
- 用一个二维数组替换了 9 个数组。
代码未经测试,请特别注意createCopy方法
// Creates each block of 9 digits copying digits from field
// row, col are the block position, starting from upper left 0, 0 to
// last block 2, 2
public static int[] createCopy(int[] field, int row, int col) {
int[] copy = new int[9];
for (int i = 3 * row; i < 3 * row + 3; i++) {
for (int j = 3 * col; j < 3 * col + 3; j++) {
copy[(i - 3 * row) * 3 + j - 3 * col] = field[i][j];
}
}
return copy;
}
// Check if one block is valid
private static boolean isValid(int[] copyArray) {
Arrays.sort(copyArray);
for (int i = 1; i < copyArray.length; i++) {
if (copyArray[i] == copyArray[i - 1]) {
// Exit immediately if not valid
return false;
}
}
return true;
}
// Create blocks, then validate them
// At first not valid block return false
public static boolean validFieldParts() {
int[][] copyArrays = new int[3][3];
for (int row = 0; row < 3; row++) {
for (int col = 0; col < 3; col++) {
copyArrays[row][col] = createCopy(field, row, col);
}
}
for (int[] copyArray : copyArrays) {
if (!isValid(copyArray)) {
// Exit immediately if not valid
return false;
}
}
return true;
}
我不会使用 9 个不同的数组和 9 个不同的循环来表示 9 的每个部分,而是使用另一个嵌套的 for 循环,它使用相同的数组迭代每个集群。
//Iterate over each 'block'
for (int row = 0; row < 3; row++) {
for (int col = 0; col < 3; col++) {
//Iterate over each cell in the block
for (int i = row*3; i < (row+1)*3; i++) {
for (int j = col*3; j < (col+1)*3; j++) {
copyArray[(i - 3) * 3 + j - 3] = field[i][j];
}
}
//Sort array and do duplication check here - return false if dupe found
}
}
return true
这会减少代码的长度,尽管它可能不会更有效率。
不使用计数器标志,而是在您递增计数器时将 return false 改为 return false 并在最后变为 return true 会快得多。这将防止不必要的代码来自 运行
我现在正在做作业,有一个关于在 Java 中重构我的代码的问题。 我现在正在玩数独,我需要检查 3x3 框是否有效。为此,我创建了一个包含所有框数的一维数组,然后比较它们的值。它现在正在工作,但实际上根本没有重构。我真的很想知道是否有任何方法可以减少所有这些复制粘贴。
public static boolean validFieldParts() {
int counter = 0;
boolean isValid = false;
int[] copyArray1 = new int[field.length];
int[] copyArray2 = new int[field.length];
int[] copyArray3 = new int[field.length];
int[] copyArray4 = new int[field.length];
int[] copyArray5 = new int[field.length];
int[] copyArray6 = new int[field.length];
int[] copyArray7 = new int[field.length];
int[] copyArray8 = new int[field.length];
int[] copyArray9 = new int[field.length];
// copy the array
// 1 große Feld
for (int i = 0; i < field.length / 3; i++) {
for (int j = 0; j < field[i].length / 3; j++) {
copyArray1[i * 3 + j] = field[i][j];
}
}
// 2 große Feld
for (int i = 0; i < field.length / 3; i++) {
for (int j = 3; j < 6; j++) {
copyArray2[i * 3 + j - 3] = field[i][j];
}
}
// 3 große Feld
for (int i = 0; i < field.length / 3; i++) {
for (int j = 6; j < 9; j++) {
copyArray3[i * 3 + j - 6] = field[i][j];
}
}
// 4 große Feld
for (int i = 3; i < 6; i++) {
for (int j = 0; j < field[i].length / 3; j++) {
copyArray4[(i - 3) * 3 + j] = field[i][j];
}
}
// 5 große Feld
for (int i = 3; i < 6; i++) {
for (int j = 3; j < 6; j++) {
copyArray5[(i - 3) * 3 + j - 3] = field[i][j];
}
}
// 6 große Feld
for (int i = 3; i < 6; i++) {
for (int j = 6; j < 9; j++) {
copyArray6[(i - 3) * 3 + j - 6] = field[i][j];
}
}
// 7 große Feld
for (int i = 6; i < 9; i++) {
for (int j = 0; j < field[i].length / 3; j++) {
copyArray7[(i - 6) * 3 + j] = field[i][j];
}
}
// 8 große Feld
for (int i = 6; i < 9; i++) {
for (int j = 3; j < 6; j++) {
copyArray8[(i - 6) * 3 + j - 3] = field[i][j];
}
}
// 9 große Feld
for (int i = 6; i < 9; i++) {
for (int j = 6; j < 9; j++) {
copyArray9[(i - 6) * 3 + j - 6] = field[i][j];
}
}
Arrays.sort(copyArray1);
Arrays.sort(copyArray2);
Arrays.sort(copyArray3);
Arrays.sort(copyArray4);
Arrays.sort(copyArray5);
Arrays.sort(copyArray6);
Arrays.sort(copyArray7);
Arrays.sort(copyArray8);
Arrays.sort(copyArray9);
for (int i = 1; i < copyArray1.length; i++) {
if (copyArray1[i] == copyArray1[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray2.length; i++) {
if (copyArray2[i] == copyArray2[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray3.length; i++) {
if (copyArray3[i] == copyArray3[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray4.length; i++) {
if (copyArray4[i] == copyArray4[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray5.length; i++) {
if (copyArray5[i] == copyArray5[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray6.length; i++) {
if (copyArray6[i] == copyArray6[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray7.length; i++) {
if (copyArray7[i] == copyArray7[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray8.length; i++) {
if (copyArray8[i] == copyArray8[i - 1])
counter++;
else
continue;
}
for (int i = 1; i < copyArray9.length; i++) {
if (copyArray9[i] == copyArray9[i - 1])
counter++;
else
continue;
}
if (counter > 0)
isValid = false;
else
isValid = true;
return isValid;
}
这里是一个完整的重构。这里的改进:
- 创建了两个新方法:
createCopy
和isValid
- 删除了未使用的变量
counter
和isValid
- 用一个二维数组替换了 9 个数组。
代码未经测试,请特别注意createCopy方法
// Creates each block of 9 digits copying digits from field
// row, col are the block position, starting from upper left 0, 0 to
// last block 2, 2
public static int[] createCopy(int[] field, int row, int col) {
int[] copy = new int[9];
for (int i = 3 * row; i < 3 * row + 3; i++) {
for (int j = 3 * col; j < 3 * col + 3; j++) {
copy[(i - 3 * row) * 3 + j - 3 * col] = field[i][j];
}
}
return copy;
}
// Check if one block is valid
private static boolean isValid(int[] copyArray) {
Arrays.sort(copyArray);
for (int i = 1; i < copyArray.length; i++) {
if (copyArray[i] == copyArray[i - 1]) {
// Exit immediately if not valid
return false;
}
}
return true;
}
// Create blocks, then validate them
// At first not valid block return false
public static boolean validFieldParts() {
int[][] copyArrays = new int[3][3];
for (int row = 0; row < 3; row++) {
for (int col = 0; col < 3; col++) {
copyArrays[row][col] = createCopy(field, row, col);
}
}
for (int[] copyArray : copyArrays) {
if (!isValid(copyArray)) {
// Exit immediately if not valid
return false;
}
}
return true;
}
我不会使用 9 个不同的数组和 9 个不同的循环来表示 9 的每个部分,而是使用另一个嵌套的 for 循环,它使用相同的数组迭代每个集群。
//Iterate over each 'block'
for (int row = 0; row < 3; row++) {
for (int col = 0; col < 3; col++) {
//Iterate over each cell in the block
for (int i = row*3; i < (row+1)*3; i++) {
for (int j = col*3; j < (col+1)*3; j++) {
copyArray[(i - 3) * 3 + j - 3] = field[i][j];
}
}
//Sort array and do duplication check here - return false if dupe found
}
}
return true
这会减少代码的长度,尽管它可能不会更有效率。
不使用计数器标志,而是在您递增计数器时将 return false 改为 return false 并在最后变为 return true 会快得多。这将防止不必要的代码来自 运行