Question

我正在研究挑战，我需要找到多米诺骨牌瓷砖线性链的所有可能性。我理解递归的原理，但不了解如何将其转换为代码。如果有人可以通过简单的步骤解释问题（解决方案），然后我可以遵循并尝试对它们进行编码。

示例：

瓷砖： [3/4] [5/6] [1/4] [1/6]

可能的链： [3/4] - [4/1] - [1/6] - [6/5]

允许翻转瓷砖。（切换号码）

Answer 1

这个过程非常简单：你从一组多米诺骨牌D和一个空链C开始。

for each domino in the collection:
    see if it can be added to the chain (either the chain is empty, or the first 
    number is the same as the second number of the last domino in the chain.
    if it can, 
        append the domino to the chain,
        then print this new chain as it is a solution,
        then call recursively with D - {domino} and C + {domino}

    repeat with the flipped domino

Java代码：

public class Domino {
    public final int a;
    public final int b;

    public Domino(int a, int b) {
        this.a = a;
        this.b = b;
    }

    public Domino flipped() {
        return new Domino(b, a);
    }

    @Override
    public String toString() {
        return "[" + a + "/" + b + "]";
    }
}

算法：

private static void listChains(List<Domino> chain, List<Domino> list) {
    for (int i = 0; i < list.size(); ++i) {
        Domino dom = list.get(i);
        if (canAppend(dom, chain)) {
            chain.add(dom);
            System.out.println(chain);
            Domino saved = list.remove(i);
            listChains(chain, list);
            list.add(i, saved);
            chain.remove(chain.size()-1);
        }
        dom = dom.flipped();
        if (canAppend(dom, chain)) {
            chain.add(dom);
            System.out.println(chain);
            Domino saved = list.remove(i);
            listChains(chain, list);
            list.add(i, saved);
            chain.remove(chain.size()-1);
        }
    }
}

private static boolean canAppend(Domino dom, List<Domino> to) {
    return to.isEmpty() || to.get(to.size()-1).b == dom.a;
}

你的例子：

public static void main(String... args) {
    List<Domino> list = new ArrayList<>();
    // [3/4] [5/6] [1/4] [1/6]
    list.add(new Domino(3, 4));
    list.add(new Domino(5, 6));
    list.add(new Domino(1, 4));
    list.add(new Domino(1, 6));

    List<Domino> chain = new ArrayList<>();
    listChains(chain, list);
}

通过递归和回溯查找所有可能的多米诺骨牌链

1 个答案: