这个过程非常简单：首先从一组多米诺骨牌D和一条空链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);
}
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