在Swift 5.3中

Question

我需要一些帮助，我正在为Universiy的Programming II课程编写一个程序。问题是要求使用递归计算Fibonacci序列。必须将计算出的斐波纳契数存储在一个数组中，以阻止不必要的重复计算并减少计算时间。

我设法让程序在没有数组和记忆的情况下工作，现在我正在尝试实现它并且我被卡住了。我不确定如何构建它。我用谷歌搜索并略读了一些书籍，但没有找到太多帮助我解决如何实施解决方案。

import javax.swing.JOptionPane;
public class question2
{
static int count = 0;
static int [] dictionary;

public static void main(String[] args)
{

int answer;
int num = Integer.parseInt(javax.swing.JOptionPane.showInputDialog("Enter n:"));

javax.swing.JOptionPane.showMessageDialog(null, 
        "About to calculate fibonacci(" + num + ")");

//giving the array "n" elements
dictionary= new int [num];

if (dictionary.length>=0)
dictionary[0]= 0;

if (dictionary.length>=1)
dictionary[0]= 0;
dictionary[1]= 1;


//method call
answer = fibonacci(num);

//output
JOptionPane.showMessageDialog(null,"Fibonacci("+num+") is "+answer+" (took "+count+" calls)");
}



  static int fibonacci(int n)
  {
count++;

// Only defined for n >= 0
if (n < 0) {
  System.out.println("ERROR: fibonacci sequence not defined for negative numbers.");
  System.exit(1);
}

// Base cases: f(0) is 0, f(1) is 1
// Other cases: f(n) = f(n-1) + f(n-2)/
if (n == 0) 
{
  return dictionary[0];
}

else if (n == 1) 
{
  return dictionary[1];
}

else
return dictionary[n] = fibonacci(n-1) + fibonacci(n-2);



}

}

以上是不正确的，我的fib方法的结束是主要问题。我不知道如何让它以递归的方式将数字添加到数组的正确部分。

Answer 1

您需要区分字典中已计算的数字和未计算的数字，而您目前不会：始终重新计算数字。

if (n == 0) 
{
  // special case because fib(0) is 0
  return dictionary[0];
}
else 
{
  int f = dictionary[n];
  if (f == 0) {
    // number wasn't calculated yet.
    f = fibonacci(n-1) + fibonacci(n-2);
    dictionary[n] = f;
  }
  return f;
}

Answer 2

public static int fib(int n, Map<Integer,Integer> map){

    if(n ==0){
        return 0;
    }

    if(n ==1){
        return 1;
    }

    if(map.containsKey(n)){
        return map.get(n);
    }

    Integer fibForN = fib(n-1,map) + fib(n-2,map);
    map.put(n, fibForN);

    return fibForN; 

}

与上述大多数解决方案类似，但使用的是地图。

Answer 3

使用Memoization打印首个n斐波纳契数字的程序。

int[] dictionary; 
// Get Fibonacci with Memoization
public int getFibWithMem(int n) {
    if (dictionary == null) {
        dictionary = new int[n];
    }

    if (dictionary[n - 1] == 0) {
        if (n <= 2) {
            dictionary[n - 1] = n - 1;
        } else {
            dictionary[n - 1] = getFibWithMem(n - 1) + getFibWithMem(n - 2);
        }
    }

    return dictionary[n - 1];
}

public void printFibonacci()
{
    for (int curr : dictionary) {
        System.out.print("F[" + i++ + "]:" + curr + ", ");
    }
}

Answer 4

我相信你忘了在字典里查找一些内容。

更改

else
    return dictionary[n] = fibonacci(n-1) + fibonacci(n-2);

到

else {
    if (dictionary[n] > 0)
        return dictionary[n];

    return dictionary[n] = fibonacci(n - 1) + fibonacci(n - 2);
}

它工作正常（我自己测试过:)：

Answer 5

这是我对递归斐波那契记忆的实现。使用BigInteger和ArrayList可以计算出第100个甚至更长的项。我尝试了第1000个术语，结果在几毫秒内返回，这里是代码：

    private static List<BigInteger> dict = new ArrayList<BigInteger>();
    public static void printFebonachiRecursion (int num){
    if (num==1){
        printFebonachiRecursion(num-1);
        System.out.printf("Term %d: %d%n",num,1);
        dict.add(BigInteger.ONE);
    }
    else if (num==0){
        System.out.printf("Term %d: %d%n",num,0);
        dict.add(BigInteger.ZERO);
    }
    else {
    printFebonachiRecursion(num-1);
    dict.add(dict.get(num-2).add(dict.get(num-1)));
    System.out.printf("Term %d: %d%n",num,dict.get(num));
    }
}

输出示例

printFebonachiRecursion(100);

Term 0: 0
Term 1: 1
Term 2: 1
Term 3: 2
...
Term 98: 135301852344706746049
Term 99: 218922995834555169026
Term 100: 354224848179261915075

Answer 6

int F(int Num){
int i =0;
int* A = NULL;
if(Num > 0)
{
 A = (int*) malloc(Num * sizeof(int));
}
else
 return Num;

for(;i<Num;i++)
 A[i] = -1;

return F_M(Num, &A);


}

int F_M(int Num, int** Ap){
int Num1 = 0;
int Num2 = 0;

if((*Ap)[Num - 1] < 0)
{
  Num1 = F_M(Num - 1, Ap);
  (*Ap)[Num -1] = Num1;
  printf("Num1:%d\n",Num1);
}
else
  Num1 = (*Ap)[Num - 1];

if((*Ap)[Num - 2] < 0)
{
  Num2 = F_M(Num - 2, Ap);
  (*Ap)[Num -2] = Num2;
  printf("Num2:%d\n",Num2);
}
else
  Num2 = (*Ap)[Num - 2];

if(0 == Num || 1 == Num)
{
 (*Ap)[Num] = Num;
 return Num;
}
else{
//  return ((*Ap)[Num - 2] > 0?(*Ap)[Num - 2] = F_M(Num -2, Ap): (*Ap)[Num - 2]  ) +     ((*Ap)[Num - 1] > 0?(*Ap)[Num - 1] = F_M(Num -1, Ap): (*Ap)[Num - 1]  );
  return (Num1 + Num2);
}

}

int main(int argc, char** argv){
int Num = 0;
if(argc>1){
sscanf(argv[1], "%d", &Num);
}

printf("F(%d) = %d", Num, F(Num));

return 0;

}

Answer 7

这是使用静态值数组来处理递归fibonacci（）方法的memoization的另一种方法 -

public static long fibArray[]=new long[50];\\Keep it as large as you need

public static long fibonacci(long n){
long fibValue=0;
if(n==0 ){
    return 0;
}else if(n==1){
    return 1;
}else if(fibArray[(int)n]!=0){
    return fibArray[(int)n];    
}
else{
    fibValue=fibonacci(n-1)+fibonacci(n-2);
    fibArray[(int) n]=fibValue;
    return fibValue;
}
}

请注意，此方法使用全局（类级别）静态数组fibArray []。要查看整个代码并附上说明，您还可以看到以下内容 - http://www.javabrahman.com/gen-java-programs/recursive-fibonacci-in-java-with-memoization/

Answer 8

这是一个利用 memoization 概念的完全成熟的课程：

import java.util.HashMap;
import java.util.Map;

public class Fibonacci {

    public static Fibonacci getInstance() {
        return new Fibonacci();
    }

    public int fib(int n) {
        HashMap<Integer, Integer> memoizedMap = new HashMap<>();

        memoizedMap.put(0, 0);
        memoizedMap.put(1, 1);

        return fib(n, memoizedMap);
    }

    private int fib(int n, Map<Integer, Integer> map) {
        if (map.containsKey(n))
            return map.get(n);

        int fibFromN = fib(n - 1, map) + fib(n - 2, map);

        // MEMOIZE the computed value
        map.put(n, fibFromN);

        return fibFromN;
    }
}

请注意

memoizedMap.put(0, 0);
memoizedMap.put(1, 1);

用于消除以下检查的必要性

if (n == 0) return 0;
if (n == 1) return 1;

在每次递归函数调用时。

Answer 9

import java.util.HashMap;
import java.util.Map;

public class FibonacciSequence {

    public static int fibonacci(int n, Map<Integer, Integer> memo) {
        if (n < 2) {
            return n;
        }
        if (!memo.containsKey(n)) {
            memo.put(n, fibonacci(n - 1, memo) + fibonacci(n - 2, memo));
        }
        return memo.get(n);
    }

    public static int fibonacci(int n, int[] memo) {
        if (n < 2) {
            return n;
        }
        if (memo[n - 1] != 0) {
            return memo[n - 1];
        }
        return memo[n - 1] = fibonacci(n - 1, memo) + fibonacci(n - 2, memo);
    }

    public static void main(String[] s) {
        int n = 10;

        System.out.println("f(n) = " + fibonacci(n, new HashMap<Integer, Integer>()));
        System.out.println("f(n) = " + fibonacci(n, new int[n]));
    }
}

Answer 10

可能太老了，但这是我的快速解决方案

class Recursion {
    func fibonacci(_ input: Int) {
        var dictioner: [Int: Int] = [:]
        dictioner[0] = 0
        dictioner[1] = 1
       print(fibonacciCal(input, dictioner: &dictioner))
    }

    func fibonacciCal(_ input: Int, dictioner: inout [Int: Int]) -> Int {
        if let va = dictioner[input]{
            return va
        } else {
            let firstPart = fibonacciCal(input-1, dictioner: &dictioner)

            let secondPart = fibonacciCal(input-2, dictioner: &dictioner)

            if dictioner[input] == nil {
                dictioner[input] = firstPart+secondPart
            }

            return firstPart+secondPart
        }
    }
}

// 0,1,1,2,3,5,8
class TestRecursion {
    func testRecursion () {
        let t = Recursion()
        t.fibonacci(3)
    }
}

Answer 11

public class FiboSeries {

    // first two terms of Fibonacci
    int x1 = 0;
    int x2 = 1;
    long xn; // nth number in Fibo series
    long[] array; // an array for implementing memoization

    // print the Nth number of Fibonacci - logic is f(n) = f(n-1) + f(n-2)

    long fibo(int n) {

        // initialize the array having n elements if it does not exist already
        if (array == null) {

            array = new long[n + 1];

        }

        // Fetch the memoized value from the array instead of recursion
        // for instance, fibo(3) will be calculated just once and stored inside this
        // array for next call
        if (array[n] != 0)

        {
            xn = array[n];
            return xn;
        }

        // value of fibo(1)
        if (n == 1) {
            xn = x1;

        }

        // value of fibo(2)
        if (n == 2) {
            xn = x2;

        }

        // value of Fibo(n) using non linear recursion
        if (n > 2) {

            xn = fibo(n - 1) + fibo(n - 2);
        }

        // before returning the value - store it at nth position of an array
        // However, before saving the value into array, check if the position is already 
        //full or not

        if (array[n] == 0) {

            array[n] = xn;
        }

        return xn;

    }

    public static void main(String[] args) {

        FiboSeries f = new FiboSeries();

        int n = 50;
        long number = f.fibo(n);

        System.out.println(number);

    }


}

Answer 12

在Swift 5.3中

这是使用备忘录的非常快速的方法。首先，我初始化缓存字典。

var cache = [Int:Int]()

然后创建我的斐波那契数字生成器。由于它是一个递归函数，因此从理论上讲，对该函数的每次调用都会再次计算整个斐波那契数列，直到请求的数量为止。这就是为什么我们使用缓存来加快递归功能的原因：

func fibonacci(_ number: Int) -> Int {
    // if the value is in the dictionary I just return it
    if let value = cache[number] { return value }

    // Otherwise I calculate it recursively. 
    // Every recursion will check the cache again, 
    // this is why memoisation is faster!
    let newValue = number < 2 ? number : fibonacci(number - 1) + fibonacci(number - 2)
    cache[number] = newValue
    return newValue
}

我可以将序列保存在这样的数组中：

var numbers = Array(0..<10).map(fibonacci) //[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

或循环使用该功能。

Answer 13

#include <stdio.h>
long int A[100]={1,1};
long int fib(int n){
    if (A[n])
    {
        return A[n];
    }
    else
    {
         return A[n]=fib(n-1)+fib(n-2);
    }
}
int main(){
    printf("%ld",fib(30));
}

Answer 14

这是我的实现。

private static int F(int N, int[] A) {
    if ((N == 0) || (N == 1)) return N;
    if (A[N] != 0) return A[N];

    if ((A[N - 1] != 0) && (A[N - 2] != 0)) {
        A[N] = A[N - 1] + A[N - 2];
        return A[N];
    }

    if (A[N-2] != 0) {
        A[N] = A[N - 2] + F(N - 1, A);
        return A[N];
    }
    if (A[N-1] != 0) {
        A[N] = A[N - 1] + F(N - 2, A);
        return A[N];
    }
    A[N] = F(N-1, A) + F(N-2, A);
    return A[N];
}

Answer 15

记忆，但采用O（log n）方法。参考http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html

private static HashMap<Integer, Integer> lookup = new HashMap<>();

private static int fib(int n){
    if(lookup.containsKey(n))
        return lookup.get(n);
    else{
        if(n <= 1)
            lookup.put(n, n);
        else{
            int k = (n & 1) == 1? (n+1)/2 : n/2;
            if((n&1) == 1)
                lookup.put(n, fib(k) * fib(k) + fib(k-1) * fib(k-1));
            else
                lookup.put(n, (2 * fib(k-1) + fib(k))*fib(k));
        }
        return lookup.get(n);
    }
}

Answer 16

使用系统；使用 System.Collections.Generic;

命名空间斐波那契 { 公共类斐波那契数列 {

        static void Main(string[] args)
    {
        int n;
        Dictionary<int, long> dict = new Dictionary<int, long>();
        Console.WriteLine("ENTER NUMBER::");
        n = Convert.ToInt32(Console.ReadLine());
        for (int j = 0; j <= n; j++)
        {
            Console.WriteLine(Fib(j, dict));
        }
    }

   

    public static long Fib(int n, Dictionary<int, long> dict)
    {
        if (n <= 1)
            return n;
        if (dict.ContainsKey(n))
            return dict[n]; 
       
        var value = Fib(n - 1,dict) + Fib(n - 2,dict);
        dict[n] = value;
        return value;
    }

}

}

递归Fibonacci memoization

16 个答案:

在Swift 5.3中