我计划在 JavaFX 中开发一个新游戏“数字形状系统”。基本上它是一个小内存游戏，其中图片与数字相关联。所以'2'='天鹅'，'5'='手(手指)'等等。因此玩家会看到练习“天鹅 + 手指 = ？”。
我想要的是遵循规则的所有可能的数学运算:
/*
* Generate all possible mathematical operations to the console with the numbers
* 0-12, where every result is (>= 0 && <= 12).
* - Mathematical operations are '+', '-', '*' and '/'.
* - The rule 'dot before line' shouldn't be used, instead the operations will
* be executed from left to right.
* - Every among result must be between (>= 0 && <= 12) and a whole number.
* - Only different numbers are allowed for the operations (an operation have
* 2 numbers). For example 2+3 is allowed, 3*3 not.
*
* A solution with recursive methods would be preferred. I want the output for
* the length 2-10.
*
* Example output with different length:
* - Length 3: 2+3(=5)*2(=10)
* - Length 5: 2+3(=5)*2(=10)+2(=12)/4(=3)
*/
我已经准备了一个示例实现，但我不知道如何将其转换为递归功能。
import java.util.ArrayList;
import java.util.List;
public class Generator {
private static final List<Double> numbers = new ArrayList<>();
private static final List<String> operations = new ArrayList<>();
static {
numbers.add(0.0);
numbers.add(1.0);
numbers.add(2.0);
numbers.add(3.0);
numbers.add(4.0);
numbers.add(5.0);
numbers.add(6.0);
numbers.add(7.0);
numbers.add(8.0);
numbers.add(9.0);
numbers.add(10.0);
numbers.add(11.0);
numbers.add(12.0);
operations.add("+");
operations.add("-");
operations.add("*");
operations.add("/");
}
private int lineCounter = 0;
public Generator() {
this.init();
}
private void init() {
}
public void generate() {
// Length 2 ###########################################################
boolean okay = false;
int lineCounter = 0;
StringBuilder sbDouble = new StringBuilder();
for (Double first : numbers) {
for (Double second : numbers) {
for (String operation : operations) {
if (first == second) {
continue;
}
if (operation.equals("/") && (first == 0.0 || second == 0.0)) {
continue;
}
double result = perform(first, operation, second);
okay = this.check(result, operation);
if (okay) {
++lineCounter;
sbDouble = new StringBuilder();
this.computeResultAsString(sbDouble, first, operation, second, result);
System.out.println(sbDouble.toString());
}
}
}
}
System.out.println("Compute with length 2: " + lineCounter + " lines");
// Length 2 ###########################################################
// Length 3 ###########################################################
okay = false;
lineCounter = 0;
sbDouble = new StringBuilder();
for (Double first : numbers) {
for (Double second : numbers) {
for (String operation1 : operations) {
if (first == second) {
continue;
}
if (operation1.equals("/") && (first == 0.0 || second == 0.0)) {
continue;
}
double result1 = perform(first, operation1, second);
okay = this.check(result1, operation1);
if (okay) {
for (Double third : numbers) {
for (String operation2 : operations) {
if (second == third) {
continue;
}
if (operation2.equals("/") && third == 0.0) {
continue;
}
double result2 = perform(result1, operation2, third);
okay = this.check(result2, operation2);
if (okay) {
++lineCounter;
sbDouble = new StringBuilder();
this.computeResultAsString(sbDouble, first, operation1, second, result1);
this.computeResultAsString(sbDouble, operation2, third, result2);
System.out.println(sbDouble.toString());
}
}
}
}
}
}
}
System.out.println("Compute with length 3: " + lineCounter + " lines");
// Length 3 ###########################################################
// Length 4 ###########################################################
okay = false;
lineCounter = 0;
sbDouble = new StringBuilder();
for (Double first : numbers) {
for (Double second : numbers) {
for (String operation1 : operations) {
if (first == second) {
continue;
}
if (operation1.equals("/") && (first == 0.0 || second == 0.0)) {
continue;
}
double result1 = perform(first, operation1, second);
okay = this.check(result1, operation1);
if (okay) {
for (Double third : numbers) {
for (String operation2 : operations) {
if (second == third) {
continue;
}
if (operation2.equals("/") && third == 0.0) {
continue;
}
double result2 = perform(result1, operation2, third);
okay = this.check(result2, operation2);
if (okay) {
for (Double forth : numbers) {
for (String operation3 : operations) {
if (third == forth) {
continue;
}
if (operation3.equals("/") && forth == 0.0) {
continue;
}
double result3 = perform(result2, operation3, forth);
okay = this.check(result3, operation3);
if (okay) {
++lineCounter;
sbDouble = new StringBuilder();
this.computeResultAsString(sbDouble, first, operation1, second, result1);
this.computeResultAsString(sbDouble, operation2, third, result2);
this.computeResultAsString(sbDouble, operation3, forth, result3);
System.out.println(sbDouble.toString());
}
}
}
}
}
}
}
}
}
}
System.out.println("Compute with length 4: " + lineCounter + " lines");
// Length 4 ###########################################################
}
private boolean check(double result, String operation) {
switch (operation) {
case "+":
case "-":
case "*": {
if (result > 0 && result <= 12) {
return true;
}
break;
}
case "/": {
if (
(Math.floor(result) == result)
&& (result >= 0 && result <= 12)
) {
return true;
}
break;
}
}
return false;
}
private double perform(double first, String operation, double second) {
double result = 0.0;
switch (operation) {
case "+": { result = first + second; break; }
case "-": { result = first - second; break; }
case "*": { result = first * second; break; }
case "/": { result = first / second; break; }
}
return result;
}
private void computeResultAsString(StringBuilder sbDouble, String operation, double second, double result) {
sbDouble.append(operation);
sbDouble.append(second);
sbDouble.append("(=");
sbDouble.append(result);
sbDouble.append(")");
}
private void computeResultAsString(StringBuilder sbDouble, double first, String operation, double second, double result) {
sbDouble.append(first);
sbDouble.append(operation);
sbDouble.append(second);
sbDouble.append("(=");
sbDouble.append(result);
sbDouble.append(")");
}
public static void main(String[] args) {
final Generator generator = new Generator();
generator.generate();
}
}
最佳答案
正如您在自己的代码中看到的，对于“长度”的每次增加，您都必须嵌套相同代码的另一个 block 。对于动态长度值，您无法做到这一点。
因此，您将代码块移动到一个方法中，并传入一个参数来表示它还需要“嵌套”多少次，即 remainingLength。然后该方法可以使用 remainingLength 的递减值调用自身，直到达到 0。
这是一个示例，使用enum作为运算符。
public static void generate(int length) {
if (length <= 0)
throw new IllegalArgumentException();
StringBuilder expr = new StringBuilder();
for (int number = 0; number <= 12; number++) {
expr.append(number);
generate(expr, number, length - 1);
expr.setLength(0);
}
}
private static void generate(StringBuilder expr, int exprTotal, int remainingLength) {
if (remainingLength == 0) {
System.out.println(expr);
return;
}
final int exprLength = expr.length();
for (int number = 0; number <= 12; number++) {
if (number != exprTotal) {
for (Operator oper : Operator.values()) {
int total = oper.method.applyAsInt(exprTotal, number);
if (total >= 0 && total <= 12) {
expr.append(oper.symbol).append(number)
.append("(=").append(total).append(")");
generate(expr, total, remainingLength - 1);
expr.setLength(exprLength);
}
}
}
}
}
private enum Operator {
PLUS ('+', Math::addExact),
MINUS ('-', Math::subtractExact),
MULTIPLY('*', Math::multiplyExact),
DIVIDE ('/', Operator::divide);
final char symbol;
final IntBinaryOperator method;
private Operator(char symbol, IntBinaryOperator method) {
this.symbol = symbol;
this.method = method;
}
private static int divide(int left, int right) {
if (right == 0 || left % right != 0)
return -1/*No exact integer value*/;
return left / right;
}
}
请注意，排列的数量增长很快:
1: 13
2: 253
3: 5,206
4: 113,298
5: 2,583,682
6: 61,064,003
7: 1,480,508,933
关于java - 使用预定义的递归创建基本数学运算，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39753988/