📅  最后修改于: 2023-12-03 15:28:44.290000             🧑  作者: Mango
本题是GATE-CS-2015(模拟测试)考试的第五道题,属于计算机科学领域的数据结构和算法题目。
你被要求实现一个门类(Gate
),每个门通过一个信号输入(0或1)打开或关闭。如果输入的信号为1,则门打开;如果输入的信号为0,则门关闭。门也有两个输出状态:打开和关闭。
门是通过两个方法 open()
和 close()
控制的。将open()
方法调用门的状态设为“打开”,将close()
方法调用门的状态设为“关闭”。
门的两个输出状态可以通过 isOpen()
和 isClosed()
方法得到。如果门打开,则 isOpen()
返回true, isClosed()
返回false;如果门关闭,则 isOpen()
返回false, isClosed()
返回true。
给定一个初始状态 initalState
(0或1),以及多个门的实例数据,实现一个门 LogicalCircuit
,该门具有以下属性:
setGate()
方法根据给定的索引值(从1开始)设置门实例对象数组中的门。
performOperation()
方法根据给定的输入信号 signal
,模拟门电路的运行,并返回门电路的结果。
Gate inputGate1 = new Gate(initialState);
Gate inputGate2 = new Gate(initialState);
Gate andGate1 = new AndGate(inputGate1, inputGate2);
Gate outputGate = new NotGate(andGate1);
LogicalCircuit circuit = new LogicalCircuit();
circuit.setGate(1, inputGate1);
circuit.setGate(2, inputGate2);
circuit.setGate(3, andGate1);
circuit.setGate(4, outputGate);
System.out.println("初始状态 = " + circuit.performOperation(1)); // false
System.out.println("非门输出状态 = " + circuit.performOperation(0)); // true
inputGate2.SetState(1);
System.out.println("切换后的状态 = " + circuit.performOperation(1)); // true
我们可以定义一个 Gate
门类,包含两个实例变量 opened
和 closed
,它们分别用于表示门是否打开和关闭;还有两个方法 open()
和 close()
,用于控制门的状态,以及两个方法 isOpen()
和 isClosed()
用于获取门的开关状态。
public class Gate {
private boolean opened;
private boolean closed;
public Gate(boolean initialState) {
if (initialState) {
open();
} else {
close();
}
}
public void open() {
opened = true;
closed = false;
}
public void close() {
opened = false;
closed = true;
}
public boolean isOpen() {
return opened;
}
public boolean isClosed() {
return closed;
}
}
在此基础上,我们可以进一步继承并实现逻辑门 AndGate
、OrGate
、NotGate
等门类,通过组合这些门类实现 LogicalCircuit
类。
public abstract class LogicGate {
protected Gate inputGate1;
protected Gate inputGate2;
public LogicGate(Gate inputGate1, Gate inputGate2) {
this.inputGate1 = inputGate1;
this.inputGate2 = inputGate2;
}
public abstract boolean performOperation();
}
public class AndGate extends LogicGate {
public AndGate(Gate inputGate1, Gate inputGate2) {
super(inputGate1, inputGate2);
}
public boolean performOperation() {
boolean i1 = inputGate1.isOpen();
boolean i2 = inputGate2.isOpen();
return i1 && i2;
}
}
public class OrGate extends LogicGate {
public OrGate(Gate inputGate1, Gate inputGate2) {
super(inputGate1, inputGate2);
}
public boolean performOperation() {
boolean i1 = inputGate1.isOpen();
boolean i2 = inputGate2.isOpen();
return i1 || i2;
}
}
public class NotGate extends LogicGate {
public NotGate(Gate inputGate) {
super(inputGate, null);
}
public boolean performOperation() {
boolean i = inputGate1.isOpen();
return !i;
}
}
public class LogicalCircuit {
private LogicGate[] circuit;
public LogicalCircuit() {
circuit = new LogicGate[4];
}
public void setGate(int index, Gate gate) {
circuit[index - 1] = gate;
}
public boolean performOperation(int signal) {
if (circuit[0] == null || circuit[1] == null || circuit[2] == null || circuit[3] == null) {
return false;
}
circuit[0].inputGate1.SetState(signal);
boolean a = circuit[2].performOperation();
circuit[3].inputGate1.SetState(a);
return circuit[3].performOperation();
}
}
最后,我们可以使用上述门类和 LogicalCircuit
类模拟门电路的运行,并得到电路的结果。