0

我有这个类来保存两个值：

public class Coord {

    public int x; 
    public int y;

    public Coord(int x, int y) { 
        this.x = x; 
        this.y = y; 
    }
}

我正在尝试在深度优先搜索算法中使用它：

x = this.move_to_x;
y = this.move_to_y;

Coord stack = new Coord(0, 0);

Stack<Coord> start = new Stack<Coord>();
Stack<Coord> visited = new Stack<Coord>();

start.push(stack);
visited.push(stack);

while (!start.empty()) {
    Coord tmp = (Coord)start.pop();

    int j,k;

    j = tmp.x;
    k = tmp.y;

    // there is only 8 possible ways to go (the neighbors)
    for (int a = -1; a < 2; a++) {
        tmp.x = j + a;

        for (int b = -1; b < 2; b++) {
            if (a == 0 && b == 0) {
                continue;
            }

            tmp.y = k + b;

            if (tmp.x < 0 || tmp.y < 0) {
                continue;
            }

            if (tmp.x > 5 || tmp.y > 5) {
                continue;
            }

            if (tmp.x == x && tmp.y == y) {
                System.out.println("end!");
                return;
            }

            Coord push = new Coord(tmp.x, tmp.y);

            System.out.println("visited: " + visited);

            if (visited.search(push) == -1) {
                System.out.println("added x " + push.x + " y " + push.y
                        + " " + visited.search(push));

                start.push(push);
                visited.push(push);
            } else {
                System.out.println("visited x " + tmp.x + " y " + tmp.y
                        + " index  " + visited.search(push));
            }
        }
    }
}

问题是该visited.search方法总是返回-1。这是日志：

visited: [Agent.ExampleAgent.Coord@1af6a711]
added x 0 y 1 -1
visited: [Agent.ExampleAgent.Coord@1af6a711, Agent.ExampleAgent.Coord@1c727896]
added x 1 y 0 -1
visited: [Agent.ExampleAgent.Coord@1af6a711, Agent.ExampleAgent.Coord@1c727896, Agent.ExampleAgent.Coord@5fbd8c6e]
added x 1 y 1 -1
visited: [Agent.ExampleAgent.Coord@1af6a711, Agent.ExampleAgent.Coord@1c727896, Agent.ExampleAgent.Coord@5fbd8c6e, Agent.ExampleAgent.Coord@427a8ba4]
added x 0 y 0 -1
visited: [Agent.ExampleAgent.Coord@1af6a711, Agent.ExampleAgent.Coord@1c727896, Agent.ExampleAgent.Coord@5fbd8c6e, Agent.ExampleAgent.Coord@427a8ba4, Agent.ExampleAgent.Coord@262f6be5]
added x 0 y 1 -1
visited: [Agent.ExampleAgent.Coord@1af6a711, Agent.ExampleAgent.Coord@1c727896, Agent.ExampleAgent.Coord@5fbd8c6e, Agent.ExampleAgent.Coord@427a8ba4, Agent.ExampleAgent.Coord@262f6be5, Agent.ExampleAgent.Coord@199d4a86]

请注意，添加到已访问堆栈的第一个元素是(0,1)，但是当搜索它时(0,1)，该方法稍后会返回-1。

4

2 回答 2

1

（我想我会发布一个答案，概述我们在聊天讨论中取得的进展）

假设类的以下状态Coord：

public class Coord {

    public int x; 
    public int y;

    public Coord(int x, int y) { 
        this.x = x; 
        this.y = y; 
    }

    @Override
    public boolean equals(Object obj){

        if (obj == null)
            return false;
        if (obj.getClass() != Coord.class)
            return false;
        if (obj == this)
            return true;

        Coord a = (Coord)obj;
        return (a.x == this.x && a.y == this.y);
    }

    @Override
    public int hashCode() {
        return x*17 + y*31;
    }

    @Override
    public String toString() {
        return "("+x+", "+y+")";
    }
}

...实现：

  • equals()因为这是堆栈搜索使用的，根据它的 Javadoc
  • hashCode()作为最佳实践，伴随equals()
  • toString()更清晰的诊断输出

这个想法是在与其余代码隔离的情况下测试堆栈搜索的功能。如果我们可以证明堆栈搜索功能正常，那么问题就出在其他地方。

证明堆栈搜索可以使用测试类来完成，例如这个：

public class CoordTest {

public static void main(String[] args) {
    System.out.println("Running tests...");

    System.out.println("Testing: equals");
    Coord c1a = new Coord(2,3);
    Coord c1b = new Coord(2,3);
    check(c1a.equals(c1b));

    System.out.println("Testing: not equals"); 
    Coord c2a = new Coord(2,3);
    Coord c2b = new Coord(6,8);
    Coord c2c = new Coord(2,8); 
    Coord c2d = new Coord(6,3);
    check(!c2a.equals(c2b));
    check(!c2a.equals(c2c));
    check(!c2a.equals(c2d));

    System.out.println("Testing: not found in empty stack"); 
    Stack<Coord> stack1 = new Stack<Coord>();
    int result1 = stack1.search(c1a);
    check(result1 == -1);

    System.out.println("Testing: not found in non-empty stack"); 
    Stack<Coord> stack2 = new Stack<Coord>();
    stack2.push(new Coord(4,5));
    stack2.push(new Coord(6,7));
    int result2 = stack2.search(c1a);
    check(result2 == -1);

    System.out.println("Testing: found in non-empty stack"); 
    Stack<Coord> stack3 = new Stack<Coord>();
    stack3.push(new Coord(4,5));
    stack3.push(new Coord(3,1));
    stack3.push(new Coord(6,7));        
    int result3 = stack3.search(new Coord(3,1));
    check(result3 == 2);        

    System.out.println("All tests completed successfully.");
}

private static void check(boolean condition) {
    if (!condition) {
        throw new RuntimeException("Condition failed!");
    }
}

}

输出：

Running tests...
Testing: equals
Testing: not equals
Testing: not found in empty stack
Testing: not found in non-empty stack
Testing: found in non-empty stack
All tests completed successfully.
于 2013-03-08T20:14:59.070 回答
1

您需要重写该equals()方法，以便true在参数的 x 和 y 等于对象的 x 和 y 值时返回。

您还应该覆盖该hashCode()方法，使其“同意”该equals()方法。

于 2013-03-08T02:49:04.340 回答