问查找所有可能的带有递归和回溯的多米诺骨牌链

EN

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

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 + "]";
    }
}

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;
}

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);
}