Java(特别是 Netbeans 和 JForms):检查二维数组每一行中的重复项
Java (specifically Netbeans and JForms): Checking duplicates in each line of 2D array
我正在 Netbeans 的 jForm/GUI 中做一个 Lotto 应用程序,它有 3 行,每行 5 个数字,我不希望每行都允许重复。第 1 行和第 3 行有一个数字是可以的,但同一行上有这些数字是不行的。
我能想到的唯一可行的方法是对其进行硬编码,最好不要那样做。
我试过:
boolean dup = false;
for (int k = 0; k < num[0].length){ //loop through columns
for (i = 0; i < num.length-1; i++) {
for (int j = i; j < inArray.length; j++){
if (num[k][i] == num[k][j]){
dup = true;
break;
}
}
}
}
还有这个:
public static boolean hasDuplicates(int [][] num) {
for (int row = 0; row < num.length; row++) {
int curRow = num[row];
Set set = Sets.newHashSet(Arrays.asList(curRow));
if (set.size() < curRow.length) {
return true;
}
}
return false;
}
我也广泛地研究了其他编码,但我找不到一个可行的编码。
我想要做的确切事情是:
通过文本字段获取用户对 Lotto 三行的输入,检查每一行是否重复,如果重复则打印到 jLabel 或将 jLabel 留空,如果没有则 运行 其余代码重复。
我目前的密码是:
private void playBtnActionPerformed(java.awt.event.ActionEvent evt) {
num[0][0] = Integer.parseInt(line00Tf.getText());
num[0][1] = Integer.parseInt(line01Tf.getText());
num[0][2] = Integer.parseInt(line02Tf.getText());
num[0][3] = Integer.parseInt(line03Tf.getText());
num[0][4] = Integer.parseInt(line04Tf.getText());
num[1][0] = Integer.parseInt(line10Tf.getText());
num[1][1] = Integer.parseInt(line11Tf.getText());
num[1][2] = Integer.parseInt(line12Tf.getText());
num[1][3] = Integer.parseInt(line13Tf.getText());
num[1][4] = Integer.parseInt(line14Tf.getText());
num[2][0] = Integer.parseInt(line20Tf.getText());
num[2][1] = Integer.parseInt(line21Tf.getText());
num[2][2] = Integer.parseInt(line22Tf.getText());
num[2][3] = Integer.parseInt(line23Tf.getText());
num[2][4] = Integer.parseInt(line24Tf.getText());
duplicateLbl.setText("");
LottoPhase1 p1 = new LottoPhase1();
p1.setNum(num);
p1.createSecret();
secret = p1.getSecret();
p1.computeCheckInput();
correctL1 = p1.getCorrectL1();
correctL2 = p1.getCorrectL2();
correctL3 = p1.getCorrectL3();
//prints secret to output
System.out.println("Phase 1 Main Secret: " + Arrays.toString(secret));
System.out.println();
displayResults0Lbl.setText(Integer.toString(secret[0]) + ", " + Integer.toString(secret[1]) + ", " + Integer.toString(secret[2]) + ", " + Integer.toString(secret[3]) + ", " + Integer.toString(secret[4]));
matched1NumLbl.setText(Integer.toString(correctL1));
matched2NumLbl.setText(Integer.toString(correctL2));
matched3NumLbl.setText(Integer.toString(correctL3));
}
第二种方法有几个错误,例如,
int curRow = num[row];
实际上应该是:
int[] curRow = num[row];
此外,您似乎正在使用 Sets,它可能来自您正在使用的某些库 (Guava、Google Common 等.)。假设您没有使用任何库,您可以将代码更改为类似于:
public static boolean hasDuplicates(int [][] num) {
for (int[] curRow : num) {
Set<Integer> set = new HashSet<>();
for (int n : curRow) {
if (!set.add(n)) {
return true;
}
}
}
return false;
}
如果您使用的是 Java 8,删除第二个 for 循环的一种方法是使用 Stream:
public static boolean hasDuplicates(int [][] num) {
for (int[] curRow : num) {
Set<Integer> set = IntStream.of(curRow).boxed().collect(Collectors.toSet());
if (set.size() < curRow.length) {
return true;
}
}
return false;
}
可以在 these 等线程中找到 Stream 的其他替代方案。
使用以下输入进行测试会产生我认为您期望的结果:
int[][] testA = {{0,1,2,3,4}, {0,1,2,3,4}, {0,1,2,3,4}}; //false
int[][] testB = {{0,1,2,3,4}, {0,2,2,3,4}, {0,1,2,3,4}}; //true
int[][] testC = {{0,1,2,3,4}, {0,1,2,3,4}, {0,4,3,3,4}}; //true
int[][] testD = {{0,1,2,3,4}, {5,6,7,8,9}, {10,11,12,13,14}}; //false
public static boolean hasDuplicates(int[][] num)
{
boolean hasDuplicate = false;
// for each line in num
for(int[] line : num)
{
// for every number in the row
for(int i = 0; i < line.length && !hasDuplicate; i++)
{
// for every number in the row
for(int j = 0; j < line.length; j++)
{
// if we are not comparing the same number
if(i != j)
{
// check for equality
if(line[i] == line[j])
{
hasDuplicate = true; // we have found a duplicate
break; // no need to keep checking; break the loop and return
}
}
}
}
}
return hasDuplicate;
}
我正在 Netbeans 的 jForm/GUI 中做一个 Lotto 应用程序,它有 3 行,每行 5 个数字,我不希望每行都允许重复。第 1 行和第 3 行有一个数字是可以的,但同一行上有这些数字是不行的。
我能想到的唯一可行的方法是对其进行硬编码,最好不要那样做。
我试过:
boolean dup = false;
for (int k = 0; k < num[0].length){ //loop through columns
for (i = 0; i < num.length-1; i++) {
for (int j = i; j < inArray.length; j++){
if (num[k][i] == num[k][j]){
dup = true;
break;
}
}
}
}
还有这个:
public static boolean hasDuplicates(int [][] num) {
for (int row = 0; row < num.length; row++) {
int curRow = num[row];
Set set = Sets.newHashSet(Arrays.asList(curRow));
if (set.size() < curRow.length) {
return true;
}
}
return false;
}
我也广泛地研究了其他编码,但我找不到一个可行的编码。
我想要做的确切事情是:
通过文本字段获取用户对 Lotto 三行的输入,检查每一行是否重复,如果重复则打印到 jLabel 或将 jLabel 留空,如果没有则 运行 其余代码重复。
我目前的密码是:
private void playBtnActionPerformed(java.awt.event.ActionEvent evt) {
num[0][0] = Integer.parseInt(line00Tf.getText());
num[0][1] = Integer.parseInt(line01Tf.getText());
num[0][2] = Integer.parseInt(line02Tf.getText());
num[0][3] = Integer.parseInt(line03Tf.getText());
num[0][4] = Integer.parseInt(line04Tf.getText());
num[1][0] = Integer.parseInt(line10Tf.getText());
num[1][1] = Integer.parseInt(line11Tf.getText());
num[1][2] = Integer.parseInt(line12Tf.getText());
num[1][3] = Integer.parseInt(line13Tf.getText());
num[1][4] = Integer.parseInt(line14Tf.getText());
num[2][0] = Integer.parseInt(line20Tf.getText());
num[2][1] = Integer.parseInt(line21Tf.getText());
num[2][2] = Integer.parseInt(line22Tf.getText());
num[2][3] = Integer.parseInt(line23Tf.getText());
num[2][4] = Integer.parseInt(line24Tf.getText());
duplicateLbl.setText("");
LottoPhase1 p1 = new LottoPhase1();
p1.setNum(num);
p1.createSecret();
secret = p1.getSecret();
p1.computeCheckInput();
correctL1 = p1.getCorrectL1();
correctL2 = p1.getCorrectL2();
correctL3 = p1.getCorrectL3();
//prints secret to output
System.out.println("Phase 1 Main Secret: " + Arrays.toString(secret));
System.out.println();
displayResults0Lbl.setText(Integer.toString(secret[0]) + ", " + Integer.toString(secret[1]) + ", " + Integer.toString(secret[2]) + ", " + Integer.toString(secret[3]) + ", " + Integer.toString(secret[4]));
matched1NumLbl.setText(Integer.toString(correctL1));
matched2NumLbl.setText(Integer.toString(correctL2));
matched3NumLbl.setText(Integer.toString(correctL3));
}
第二种方法有几个错误,例如,
int curRow = num[row];
实际上应该是:
int[] curRow = num[row];
此外,您似乎正在使用 Sets,它可能来自您正在使用的某些库 (Guava、Google Common 等.)。假设您没有使用任何库,您可以将代码更改为类似于:
public static boolean hasDuplicates(int [][] num) {
for (int[] curRow : num) {
Set<Integer> set = new HashSet<>();
for (int n : curRow) {
if (!set.add(n)) {
return true;
}
}
}
return false;
}
如果您使用的是 Java 8,删除第二个 for 循环的一种方法是使用 Stream:
public static boolean hasDuplicates(int [][] num) {
for (int[] curRow : num) {
Set<Integer> set = IntStream.of(curRow).boxed().collect(Collectors.toSet());
if (set.size() < curRow.length) {
return true;
}
}
return false;
}
可以在 these 等线程中找到 Stream 的其他替代方案。 使用以下输入进行测试会产生我认为您期望的结果:
int[][] testA = {{0,1,2,3,4}, {0,1,2,3,4}, {0,1,2,3,4}}; //false
int[][] testB = {{0,1,2,3,4}, {0,2,2,3,4}, {0,1,2,3,4}}; //true
int[][] testC = {{0,1,2,3,4}, {0,1,2,3,4}, {0,4,3,3,4}}; //true
int[][] testD = {{0,1,2,3,4}, {5,6,7,8,9}, {10,11,12,13,14}}; //false
public static boolean hasDuplicates(int[][] num)
{
boolean hasDuplicate = false;
// for each line in num
for(int[] line : num)
{
// for every number in the row
for(int i = 0; i < line.length && !hasDuplicate; i++)
{
// for every number in the row
for(int j = 0; j < line.length; j++)
{
// if we are not comparing the same number
if(i != j)
{
// check for equality
if(line[i] == line[j])
{
hasDuplicate = true; // we have found a duplicate
break; // no need to keep checking; break the loop and return
}
}
}
}
}
return hasDuplicate;
}