将递归实现转换为基于循环的实现

Question

我有两个接口负责持有一个闭包

这是第一个在映射操作时保持闭包的方法.

package com.fs;

/**
 * This interface is responsible for holding the closures when it comes to map.
 * It uses two generic types. One for the argument and one for the return type.
 * @param <B> Generic type 
 * @param <A> Generic type
 */
public interface Func<B,A> {
    /**
     * Function prototype m takes an argument of type A and returns a type B.
     * A map operation can produce a different type.
     * @param x of type A
     * @return type B
     */
     B m(A x); 
}

第二个用于过滤操作

package com.fs;


/**
 * This interface is responsible for holding the closures when it comes to filter.
 * @param <A> Generic type 
 */
public interface Pred<A> {
    /**
     * Function prototype m takes an argument of type A and returns a boolean.
     * A filter operation checks every element if it fits a predicate.
     * @param x of type A
     * @return boolean
     */
    boolean m(A x);
}

我有一个名为CList的类,它能够使用闭包.

package com.impl.list;

import com.fs.*;

public class CList<T> {
    T head;
    CList<T> tail;

    public CList(T x, CList<T> xs){
        head = x;
        tail = xs;
    }

    static <A,B> CList<B> map(Func<B,A> f, CList<A> xs){
        if(xs==null){
            return null;
        }
        return new CList<>(f.m(xs.head),map(f,xs.tail));
    }

    static <A,B> CList<B> maploop(Func<B,A> f, CList<A> xs){
        //?????
        return null;
    }

    static <A> CList<A> filter(Pred<A> f, CList<A> xs){
        if(xs == null){
            return null;
        }
        if(f.m(xs.head)){
            return new CList<>(xs.head, filter(f,xs.tail));
        }
        return filter(f,xs.tail);
    }

    static <A> int length(CList<A> xs){
        int ans =0;
        while(xs!= null){
            ++ans;
            xs=xs.tail;
        }
        return ans;
    }
}

这是我使用闭包实现CList的公共接口.

package com.impl.list;

import com.fs.Func;
import com.fs.Pred;

public class CListClient {
    public static CList<Integer> doubleAll(CList<Integer> xs){
        Func<Integer, Integer> df = new Func<Integer, Integer>() {
            @Override
            public Integer m(Integer x) {
                return x * 2;
            }
        };

        return CList.map(df, xs);
    }

    public static int countNs(CList<Integer> xs,final int n){
        Pred<Integer> pf = new Pred<Integer>() {
            @Override
            public boolean m(Integer x) {
                return x==n;
            }
        };
        return CList.length(CList.filter(pf, xs));
    }

    public static CList<Integer> doubleAllloop(CList<Integer> xs){
        Func<Integer, Integer> df = new Func<Integer, Integer>() {
            @Override
            public Integer m(Integer x) {
                return x * 2;
            }
        };

        return CList.maploop(df, xs);
    }
}

一个简单的测试人员:

package basic;


import com.impl.list.CList;
import com.impl.list.CListClient;
import org.junit.Test;


public class ListTester {

    CList<Integer> intlist_1 = new CList<>(new Integer(1),null);
    CList<Integer> intlist_2 = new CList<>(new Integer(2),intlist_1);
    CList<Integer> intlist_3 = new CList<>(new Integer(3),intlist_2);
    CList<Integer> intlist_4 = new CList<>(new Integer(4),intlist_3);
    CList<Integer> intlist_5 = new CList<>(new Integer(4),intlist_4);
    CList<Integer> intlist = new CList<>(new Integer(5),intlist_5);

    @Test
    public void test_doubleAll(){
        CList<Integer> doubled = CListClient.doubleAll(intlist);
        CList<Integer> doubledloop = CListClient.doubleAllloop(intlist);

    }

    @Test
    public void test_CountNs(){
        int count3s = CListClient.countNs(intlist, 3);
    }
}

我试图将以递归方式实现的map函数转换为while循环.我把它命名为maploop.它让我的大脑受伤了两天.任何提示都会让我非常高兴.我在这里问这个问题,因为有可能有人可能会参加Dan Grossman课程并看到这个例子并试图实现这个功能.我更喜欢暗示而非实际答案.谢谢.

Answer 1

将递归函数转换为迭代函数时,必须检查需要哪种非尾调用状态(如果有).然后创建一个堆栈并将状态推送到堆栈,并像编码一样编写它,否则就像递归函数一样.如果函数中有多个递归调用,则需要新的状态项还包含一个值,指示函数中的哪个点.

在这种情况下,您只有一个递归调用,唯一的状态是xs,所以事情非常简单,您不需要自定义状态对象.

这是我做事的方式(未经测试).

static <A,B> CList<B> maploop(Func<B,A> f, CList<A> xs){
    Stack<CList<A>> stack = new Stack<>();

    while(xs != null){
        stack.push(xs);
        xs = xs.tail;
    }

    CList<a> result = xs;

    while(!stack.empty()){
        xs = stack.pop();
        result = new CList<>(f.m(xs.head), result);
    }

    return result;
}