ML /（Strict）Haskell Java
这是来自实际的实际项目。它利用了持久不变的数据结构，即使没有必要也使用递归。实际上，它更像Java中的Kore（项目实现的语言），但是样式基本上与ML相同。但是Kore的哲学是作者不应该格式化他的代码，因此也不格式化Java代码（它会被eclipse自动格式化）。
从列表中删除n个元素：
public static <T> List<T> drop(List<T> l, Integer n) {
return n == 0 ? l : drop(l.cons().tail, n - 1);
}
在ML / Haskell中，您要进行模式匹配以提取头部和尾部，在这里您说list.cons().x和list.cons().tail。
在列表中插入一个元素：
public static <T> List<T> insert(List<T> l, Integer i, T x) {
if (i == 0)
return cons(x, l);
return cons(l.cons().x, insert(l.cons().tail, i - 1, x));
}
从字面上定义 List是如何定义代数数据类型的。这是删除了eclipse生成的样板的版本：
public final class List<T> {
public static final class Nil<T> {
}
public static final class Cons<T> {
public final T x;
public final List<T> tail;
public Cons(T x, List<T> tail) {
if (x == null)
throw new RuntimeException("null head");
if (tail == null)
throw new RuntimeException("null tail");
this.x = x;
this.tail = tail;
}
}
private final Nil<T> nil;
private final Cons<T> cons;
private List(Nil<T> nil, Cons<T> cons) {
this.nil = nil;
this.cons = cons;
}
public boolean isEmpty() {
return nil != null;
}
public Nil<T> nil() {
if (nil == null)
throw new RuntimeException("not nil");
return nil;
}
public Cons<T> cons() {
if (cons == null)
throw new RuntimeException("not cons");
return cons;
}
public static <T> List<T> cons(Cons<T> cons) {
if (cons == null)
throw new RuntimeException("constructor received null");
return new List<T>(null, cons);
}
public static <T> List<T> nil(Nil<T> nil) {
if (nil == null)
throw new RuntimeException("constructor received null");
return new List<T>(nil, null);
}
}
这是根据特里实现的地图数据结构：
public final class Map<K, V> {
private final Tree<Character, Optional<Pair<K, V>>> tree;
// keys are sorted in reverse order so entrySet can use cons instead of append
private final Comparer<Pair<Character, Tree<Character, Optional<Pair<K, V>>>>> comparer =
new PairLeftComparer<Character, Tree<Character, Optional<Pair<K, V>>>>(
new ReverseComparer<Character>(new CharacterComparer()));
private Map(Tree<Character, Optional<Pair<K, V>>> tree) {
this.tree = tree;
}
public static <K, V> Map<K, V> empty() {
return new Map<K, V>(new Tree<Character, Optional<Pair<K, V>>>(
OptionalUtils.<Pair<K, V>> nothing(),
ListUtils
.<Pair<Character, Tree<Character, Optional<Pair<K, V>>>>> nil()));
}
public Optional<V> get(K k) {
Tree<Character, Optional<Pair<K, V>>> t = tree;
for (char c : k.toString().toCharArray()) {
Tree<Character, Optional<Pair<K, V>>> t2 = getEdge(t, c);
if (t2 == null)
return nothing();
t = t2;
}
if (t.v.isNothing())
return nothing();
return some(t.v.some().x.y);
}
public Map<K, V> put(K k, V v) {
return new Map<K, V>(put(tree, k.toString(), v, k));
}
private Tree<Character, Optional<Pair<K, V>>> put(
Tree<Character, Optional<Pair<K, V>>> t, String s, V v, K k) {
if (s.equals(""))
return new Tree<Character, Optional<Pair<K, V>>>(some(Pair.pair(k, v)),
t.edges);
char c = s.charAt(0);
Tree<Character, Optional<Pair<K, V>>> t2 = getEdge(t, c);
if (t2 == null)
return new Tree<Character, Optional<Pair<K, V>>>(
t.v,
sort(
cons(
pair(
c,
put(new Tree<Character, Optional<Pair<K, V>>>(
OptionalUtils.<Pair<K, V>> nothing(),
ListUtils
.<Pair<Character, Tree<Character, Optional<Pair<K, V>>>>> nil()),
s.substring(1), v, k)), t.edges), comparer));
return new Tree<Character, Optional<Pair<K, V>>>(t.v, sort(
replace(pair(c, put(t2, s.substring(1), v, k)), t.edges), comparer));
}
private List<Pair<Character, Tree<Character, Optional<Pair<K, V>>>>> replace(
Pair<Character, Tree<Character, Optional<Pair<K, V>>>> edge,
List<Pair<Character, Tree<Character, Optional<Pair<K, V>>>>> edges) {
if (edges.cons().x.x.equals(edge.x))
return cons(edge, edges.cons().tail);
return cons(edges.cons().x, replace(edge, edges.cons().tail));
}
// I consider this O(1). There are a constant of 2^16 values of
// char. Either way it's unusual to have a large amount of
// edges since only ASCII chars are typically used.
private Tree<Character, Optional<Pair<K, V>>> getEdge(
Tree<Character, Optional<Pair<K, V>>> t, char c) {
for (Pair<Character, Tree<Character, Optional<Pair<K, V>>>> p : iter(t.edges))
if (p.x.equals(c))
return p.y;
return null;
}
public Map<K, V> delete(K k) {
return new Map<K, V>(delete(tree, k.toString()).x);
}
private Pair<Tree<Character, Optional<Pair<K, V>>>, Boolean> delete(
Tree<Character, Optional<Pair<K, V>>> t, String k) {
if (k.equals(""))
return pair(
new Tree<Character, Optional<Pair<K, V>>>(
OptionalUtils.<Pair<K, V>> nothing(), t.edges), t.edges.isEmpty());
char c = k.charAt(0);
Tree<Character, Optional<Pair<K, V>>> t2 = getEdge(t, c);
if (t2 == null)
return pair(t, false);
Pair<Tree<Character, Optional<Pair<K, V>>>, Boolean> p =
delete(t2, k.substring(1));
List<Pair<Character, Tree<Character, Optional<Pair<K, V>>>>> edges = nil();
for (Pair<Character, Tree<Character, Optional<Pair<K, V>>>> e : iter(t.edges))
if (!e.x.equals(c))
edges = cons(e, edges);
if (!p.y)
return pair(
new Tree<Character, Optional<Pair<K, V>>>(t.v, cons(pair(c, p.x),
edges)), false);
boolean oneEdge = t.edges.cons().tail.isEmpty();
return pair(new Tree<Character, Optional<Pair<K, V>>>(t.v, edges), oneEdge
&& t.v.isNothing());
}
public static class Entry<K, V> {
public Entry(K k, V v) {
this.k = k;
this.v = v;
}
public final K k;
public final V v;
}
public List<Entry<K, V>> entrySet() {
return entrySet(ListUtils.<Entry<K, V>> nil(), tree);
}
private List<Entry<K, V>> entrySet(List<Entry<K, V>> l,
Tree<Character, Optional<Pair<K, V>>> t) {
if (!t.v.isNothing()) {
Pair<K, V> p = t.v.some().x;
l = cons(new Entry<K, V>(p.x, p.y), l);
}
for (Pair<Character, Tree<Character, Optional<Pair<K, V>>>> e : iter(t.edges))
l = entrySet(l, e.y);
return l;
}
}
这些类型开始占用与代码一样多的空间。例如，在put中，该方法具有302个类型的字符和343个代码的字符（不计算空格/换行符）。
.litcoffee。这可能会有所帮助。