问题
Please explain this simple code:
public int fibonacci(int n) {
if(n == 0)
return 0;
else if(n == 1)
return 1;
else
return fibonacci(n - 1) + fibonacci(n - 2);
}
I\'m confused with the last line especially because if n = 5 for example, then fibonacci(4) + fibonacci(3) would be called and so on but I don\'t understand how this algorithm calculates the value at index 5 by this method. Please explain with a lot of detail!
回答1:
In fibonacci sequence each item is the sum of the previous two. So, you wrote a recursive algorithm.
So,
fibonacci(5) = fibonacci(4) + fibonacci(3)
fibonacci(3) = fibonacci(2) + fibonacci(1)
fibonacci(4) = fibonacci(3) + fibonacci(2)
fibonacci(2) = fibonacci(1) + fibonacci(0)
Now you already know fibonacci(1)==1 and fibonacci(0) == 0. So, you can subsequently calculate the other values.
Now,
fibonacci(2) = 1+0 = 1
fibonacci(3) = 1+1 = 2
fibonacci(4) = 2+1 = 3
fibonacci(5) = 3+2 = 5
And from fibonacci sequence 0,1,1,2,3,5,8,13,21.... we can see that for 5th element the fibonacci sequence returns 5.
See here for Recursion Tutorial.
回答2:
There are 2 issues with your code:
- The result is stored in int which can handle only a first 48 fibonacci numbers, after this the integer fill minus bit and result is wrong.
- But you never can run fibonacci(50).
The codefibonacci(n - 1) + fibonacci(n - 2)
is very wrong.
The problem is that the it calls fibonacci not 50 times but much more.
At first it calls fibonacci(49)+fibonacci(48),
next fibonacci(48)+fibonacci(47) and fibonacci(47)+fibonacci(46)
Each time it became fibonacci(n) worse, so the complexity is exponential.
The approach to non-recursive code:
double fibbonaci(int n){
double prev=0d, next=1d, result=0d;
for (int i = 0; i < n; i++) {
result=prev+next;
prev=next;
next=result;
}
return result;
}
回答3:
In pseudo code, where n = 5, the following takes place:
fibonacci(4) + fibonnacci(3)
This breaks down into:
(fibonacci(3) + fibonnacci(2)) + (fibonacci(2) + fibonnacci(1))
This breaks down into:
(((fibonacci(2) + fibonnacci(1)) + ((fibonacci(1) + fibonnacci(0))) + (((fibonacci(1) + fibonnacci(0)) + 1))
This breaks down into:
((((fibonacci(1) + fibonnacci(0)) + 1) + ((1 + 0)) + ((1 + 0) + 1))
This breaks down into:
((((1 + 0) + 1) + ((1 + 0)) + ((1 + 0) + 1))
This results in: 5
Given the fibonnacci sequence is 1 1 2 3 5 8 ..., the 5th element is 5. You can use the same methodology to figure out the other iterations.
回答4:
Recursion can be hard to grasp sometimes. Just evaluate it on a piece of paper for a small number:
fib(4)
-> fib(3) + fib(2)
-> fib(2) + fib(1) + fib(1) + fib(0)
-> fib(1) + fib(0) + fib(1) + fib(1) + fib(0)
-> 1 + 0 + 1 + 1 + 0
-> 3
I am not sure how Java actually evaluates this, but the result will be the same.
回答5:
You can also simplify your function, as follows:
public int fibonacci(int n) {
if (n < 2) return n;
return fibonacci(n - 1) + fibonacci(n - 2);
}
回答6:
F(n)
/ \
F(n-1) F(n-2)
/ \ / \
F(n-2) F(n-3) F(n-3) F(n-4)
/ \
F(n-3) F(n-4)
Important point to note is this algorithm is exponential because it does not store the result of previous calculated numbers. eg F(n-3) is called 3 times.
For more details refer algorithm by dasgupta chapter 0.2
回答7:
Most of the answers are good and explains how the recursion in fibonacci works.
Here is an analysis on the three techniques which includes recursion as well:
- For Loop
- Recursion
- Memoization
Here is my code to test all three:
public class Fibonnaci {
// Output = 0 1 1 2 3 5 8 13
static int fibMemo[];
public static void main(String args[]) {
int num = 20;
System.out.println("By For Loop");
Long startTimeForLoop = System.nanoTime();
// returns the fib series
int fibSeries[] = fib(num);
for (int i = 0; i < fibSeries.length; i++) {
System.out.print(" " + fibSeries[i] + " ");
}
Long stopTimeForLoop = System.nanoTime();
System.out.println("");
System.out.println("For Loop Time:" + (stopTimeForLoop - startTimeForLoop));
System.out.println("By Using Recursion");
Long startTimeRecursion = System.nanoTime();
// uses recursion
int fibSeriesRec[] = fibByRec(num);
for (int i = 0; i < fibSeriesRec.length; i++) {
System.out.print(" " + fibSeriesRec[i] + " ");
}
Long stopTimeRecursion = System.nanoTime();
System.out.println("");
System.out.println("Recursion Time:" + (stopTimeRecursion -startTimeRecursion));
System.out.println("By Using Memoization Technique");
Long startTimeMemo = System.nanoTime();
// uses memoization
fibMemo = new int[num];
fibByRecMemo(num-1);
for (int i = 0; i < fibMemo.length; i++) {
System.out.print(" " + fibMemo[i] + " ");
}
Long stopTimeMemo = System.nanoTime();
System.out.println("");
System.out.println("Memoization Time:" + (stopTimeMemo - startTimeMemo));
}
//fib by memoization
public static int fibByRecMemo(int num){
if(num == 0){
fibMemo[0] = 0;
return 0;
}
if(num ==1 || num ==2){
fibMemo[num] = 1;
return 1;
}
if(fibMemo[num] == 0){
fibMemo[num] = fibByRecMemo(num-1) + fibByRecMemo(num -2);
return fibMemo[num];
}else{
return fibMemo[num];
}
}
public static int[] fibByRec(int num) {
int fib[] = new int[num];
for (int i = 0; i < num; i++) {
fib[i] = fibRec(i);
}
return fib;
}
public static int fibRec(int num) {
if (num == 0) {
return 0;
} else if (num == 1 || num == 2) {
return 1;
} else {
return fibRec(num - 1) + fibRec(num - 2);
}
}
public static int[] fib(int num) {
int fibSum[] = new int[num];
for (int i = 0; i < num; i++) {
if (i == 0) {
fibSum[i] = i;
continue;
}
if (i == 1 || i == 2) {
fibSum[i] = 1;
continue;
}
fibSum[i] = fibSum[i - 1] + fibSum[i - 2];
}
return fibSum;
}
}
Here are the results:
By For Loop
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181
For Loop Time:347688
By Using Recursion
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181
Recursion Time:767004
By Using Memoization Technique
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181
Memoization Time:327031
Hence we can see memoization is the best time wise and for loop matches closely.
But recursion takes the longest and may be you should avoid in real life. Also if you are using recursion make sure that you optimize the solution.
回答8:
This is the best video I have found that fully explains recursion and the Fibonacci sequence in Java.
http://www.youtube.com/watch?v=dsmBRUCzS7k
This is his code for the sequence and his explanation is better than I could ever do trying to type it out.
public static void main(String[] args)
{
int index = 0;
while (true)
{
System.out.println(fibonacci(index));
index++;
}
}
public static long fibonacci (int i)
{
if (i == 0) return 0;
if (i<= 2) return 1;
long fibTerm = fibonacci(i - 1) + fibonacci(i - 2);
return fibTerm;
}
回答9:
For fibonacci recursive solution, it is important to save the output of smaller fibonacci numbers, while retrieving the value of larger number. This is called "Memoizing".
Here is a code that use memoizing the smaller fibonacci values, while retrieving larger fibonacci number. This code is efficient and doesn't make multiple requests of same function.
import java.util.HashMap;
public class Fibonacci {
private HashMap<Integer, Integer> map;
public Fibonacci() {
map = new HashMap<>();
}
public int findFibonacciValue(int number) {
if (number == 0 || number == 1) {
return number;
}
else if (map.containsKey(number)) {
return map.get(number);
}
else {
int fibonacciValue = findFibonacciValue(number - 2) + findFibonacciValue(number - 1);
map.put(number, fibonacciValue);
return fibonacciValue;
}
}
}
回答10:
in the fibonacci sequence, the first two items are 0 and 1, each other item is the sum of the two previous items. i.e:
0 1 1 2 3 5 8...
so the 5th item is the sum of the 4th and the 3rd items.
回答11:
Michael Goodrich et al provide a really clever algorithm in Data Structures and Algorithms in Java, for solving fibonacci recursively in linear time by returning an array of [fib(n), fib(n-1)].
public static long[] fibGood(int n) {
if (n < = 1) {
long[] answer = {n,0};
return answer;
} else {
long[] tmp = fibGood(n-1);
long[] answer = {tmp[0] + tmp[1], tmp[0]};
return answer;
}
}
This yields fib(n) = fibGood(n)[0].
回答12:
A Fibbonacci sequence is one that sums the result of a number when added to the previous result starting with 1.
so.. 1 + 1 = 2
2 + 3 = 5
3 + 5 = 8
5 + 8 = 13
8 + 13 = 21
Once we understand what Fibbonacci is, we can begin to break down the code.
public int fibonacci(int n) {
if(n == 0)
return 0;
else if(n == 1)
return 1;
else
return fibonacci(n - 1) + fibonacci(n - 2);
}
The first if statment checks for a base case, where the loop can break out. The else if statement below that is doing the same, but it could be re-written like so...
public int fibonacci(int n) {
if(n < 2)
return n;
return fibonacci(n - 1) + fibonacci(n - 2);
}
Now that a base case is establish we have to understand the call stack.Your first call to "fibonacci" will be the last to resolve on the stack (sequence of calls) as they resolve in the reverse order from which they were called. The last method called resolves first, then the last to be called before that one and so on...
So, all the calls are made first before anything is "calculated" with those results. With an input of 8 we expect an output of 21 (see table above).
fibonacci(n - 1) keeps being called until it reaches the base case, then fibonacci(n - 2) is called until it reaches the base case. When the stack starts summing the result in reverse order, the result will be like so...
1 + 1 = 1 ---- last call of the stack (hits a base case).
2 + 1 = 3 ---- Next level of the stack (resolving backwards).
2 + 3 = 5 ---- Next level of the stack (continuing to resolve).
They keep bubbling (resolving backwards) up until the correct sum is returned to the first call in the stack and that's how you get your answer.
Having said that, this algorithm is very inefficient because it calculates the same result for each branch the code splits into. A much better approach is a "bottom up" one where no Memoization (caching) or recursion (deep call stack) is required.
Like so...
static int BottomUpFib(int current)
{
if (current < 2) return current;
int fib = 1;
int last = 1;
for (int i = 2; i < current; i++)
{
int temp = fib;
fib += last;
last = temp;
}
return fib;
}
回答13:
Most of solutions offered here run in O(2^n) complexity. Recalculating identical nodes in recursive tree is inefficient and wastes CPU cycles.
We can use memoization to make fibonacci function run in O(n) time
public static int fibonacci(int n) {
return fibonacci(n, new int[n + 1]);
}
public static int fibonacci(int i, int[] memo) {
if (i == 0 || i == 1) {
return i;
}
if (memo[i] == 0) {
memo[i] = fibonacci(i - 1, memo) + fibonacci(i - 2, memo);
}
return memo[i];
}
If we follow Bottom-Up Dynamic Programming route, below code is simple enough to compute fibonacci:
public static int fibonacci1(int n) {
if (n == 0) {
return n;
} else if (n == 1) {
return n;
}
final int[] memo = new int[n];
memo[0] = 0;
memo[1] = 1;
for (int i = 2; i < n; i++) {
memo[i] = memo[i - 1] + memo[i - 2];
}
return memo[n - 1] + memo[n - 2];
}
回答14:
Why this answer is different
Every other answer either:
- Prints instead of returns
- Makes 2 recursive calls per iteration
- Ignores the question by using loops
(aside: none of these is actually efficient; use Binet's formula to directly calculate the nth term)
Tail Recursive Fib
Here is a recursive approach that avoids a double-recursive call by passing both the previous answer AND the one before that.
private static final int FIB_0 = 0;
private static final int FIB_1 = 1;
private int calcFibonacci(final int target) {
if (target == 0) { return FIB_0; }
if (target == 1) { return FIB_1; }
return calcFibonacci(target, 1, FIB_1, FIB_0);
}
private int calcFibonacci(final int target, final int previous, final int fibPrevious, final int fibPreviousMinusOne) {
final int current = previous + 1;
final int fibCurrent = fibPrevious + fibPreviousMinusOne;
// If you want, print here / memoize for future calls
if (target == current) { return fibCurrent; }
return calcFibonacci(target, current, fibCurrent, fibPrevious);
}
回答15:
It is a basic sequence that display or get a output of 1 1 2 3 5 8 it is a sequence that the sum of previous number the current number will be display next.
Try to watch link below Java Recursive Fibonacci sequence Tutorial
public static long getFibonacci(int number){
if(number<=1) return number;
else return getFibonacci(number-1) + getFibonacci(number-2);
}
Click Here Watch Java Recursive Fibonacci sequence Tutorial for spoon feeding
回答16:
I think this is a simple way:
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int number = input.nextInt();
long a = 0;
long b = 1;
for(int i = 1; i<number;i++){
long c = a +b;
a=b;
b=c;
System.out.println(c);
}
}
}
回答17:
RanRag(accepted) answer will work fine but that's not optimized solution until and unless it is memorized as explained in Anil answer.
For recursive consider below approach, method calls of TestFibonacci are minimum
public class TestFibonacci {
public static void main(String[] args) {
int n = 10;
if (n == 1) {
System.out.println(1);
} else if (n == 2) {
System.out.println(1);
System.out.println(1);
} else {
System.out.println(1);
System.out.println(1);
int currentNo = 3;
calFibRec(n, 1, 1, currentNo);
}
}
public static void calFibRec(int n, int secondLast, int last,
int currentNo) {
if (currentNo <= n) {
int sum = secondLast + last;
System.out.println(sum);
calFibRec(n, last, sum, ++currentNo);
}
}
}
回答18:
By using an internal ConcurrentHashMap which theoretically might allow this recursive implementation to properly operate in a multithreaded environment, I have implemented a fib function that uses both BigInteger and Recursion. Takes about 53ms to calculate the first 100 fib numbers.
private final Map<BigInteger,BigInteger> cacheBig
= new ConcurrentHashMap<>();
public BigInteger fibRecursiveBigCache(BigInteger n) {
BigInteger a = cacheBig.computeIfAbsent(n, this::fibBigCache);
return a;
}
public BigInteger fibBigCache(BigInteger n) {
if ( n.compareTo(BigInteger.ONE ) <= 0 ){
return n;
} else if (cacheBig.containsKey(n)){
return cacheBig.get(n);
} else {
return
fibBigCache(n.subtract(BigInteger.ONE))
.add(fibBigCache(n.subtract(TWO)));
}
}
The test code is:
@Test
public void testFibRecursiveBigIntegerCache() {
long start = System.currentTimeMillis();
FibonacciSeries fib = new FibonacciSeries();
IntStream.rangeClosed(0,100).forEach(p -&R {
BigInteger n = BigInteger.valueOf(p);
n = fib.fibRecursiveBigCache(n);
System.out.println(String.format("fib of %d is %d", p,n));
});
long end = System.currentTimeMillis();
System.out.println("elapsed:" +
(end - start) + "," +
((end - start)/1000));
}
and output from the test is:
.
.
.
.
.
fib of 93 is 12200160415121876738
fib of 94 is 19740274219868223167
fib of 95 is 31940434634990099905
fib of 96 is 51680708854858323072
fib of 97 is 83621143489848422977
fib of 98 is 135301852344706746049
fib of 99 is 218922995834555169026
fib of 100 is 354224848179261915075
elapsed:58,0
回答19:
Here is a one line febonacci recursive:
public long fib( long n ) {
return n <= 0 ? 0 : n == 1 ? 1 : fib( n - 1 ) + fib( n - 2 );
}
回答20:
Try this
private static int fibonacci(int n){
if(n <= 1)
return n;
return fibonacci(n - 1) + fibonacci(n - 2);
}
For more info checkout thisOutput Fibonacci Series in Java - Mediocre Code
回答21:
Just to complement, if you want to be able to calculate larger numbers, you should use BigInteger.
An iterative example.
import java.math.BigInteger;
class Fibonacci{
public static void main(String args[]){
int n=10000;
BigInteger[] vec = new BigInteger[n];
vec[0]=BigInteger.ZERO;
vec[1]=BigInteger.ONE;
// calculating
for(int i = 2 ; i<n ; i++){
vec[i]=vec[i-1].add(vec[i-2]);
}
// printing
for(int i = vec.length-1 ; i>=0 ; i--){
System.out.println(vec[i]);
System.out.println("");
}
}
}
回答22:
http://en.wikipedia.org/wiki/Fibonacci_number in more details
public class Fibonacci {
public static long fib(int n) {
if (n <= 1) return n;
else return fib(n-1) + fib(n-2);
}
public static void main(String[] args) {
int N = Integer.parseInt(args[0]);
for (int i = 1; i <= N; i++)
System.out.println(i + ": " + fib(i));
}
}
Make it that as simple as needed no need to use while loop and other loop
回答23:
public class febo
{
public static void main(String...a)
{
int x[]=new int[15];
x[0]=0;
x[1]=1;
for(int i=2;i<x.length;i++)
{
x[i]=x[i-1]+x[i-2];
}
for(int i=0;i<x.length;i++)
{
System.out.println(x[i]);
}
}
}
回答24:
public class FibonacciSeries {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int N = scanner.nextInt();
for (int i = 0; i <= N; i++) {
int result = fibonacciSeries(i);
System.out.println(result);
}
scanner.close();
}
private static int fibonacciSeries(int n) {
if (n < 0) {
return 1;
} else if (n > 0) {
return fibonacciSeries(n - 1) + fibonacciSeries(n - 2);
}
return 0;
}
}
回答25:
Use while:
public int fib(int index) {
int tmp = 0, step1 = 0, step2 = 1, fibNumber = 0;
while (tmp < index - 1) {
fibNumber = step1 + step2;
step1 = step2;
step2 = fibNumber;
tmp += 1;
};
return fibNumber;
}
The advantage of this solution is that it's easy to read the code and understand it, hoping it helps
回答26:
A Fibbonacci sequence is one that sums the result of a number then we have added to the previous result, we should started from 1. I was trying to find a solution based on algorithm, so i build the recursive code, noticed that i keep the previous number and i changed the position. I'm searching the Fibbonacci sequence from 1 to 15.
public static void main(String args[]) {
numbers(1,1,15);
}
public static int numbers(int a, int temp, int target)
{
if(target <= a)
{
return a;
}
System.out.print(a + " ");
a = temp + a;
return numbers(temp,a,target);
}
回答27:
Here is O(1) solution :
private static long fibonacci(int n) {
double pha = pow(1 + sqrt(5), n);
double phb = pow(1 - sqrt(5), n);
double div = pow(2, n) * sqrt(5);
return (long) ((pha - phb) / div);
}
Binet's Fibonacci number formula used for above implementation.
For large inputs long can be replaced with BigDecimal.
回答28:
public static long fib(int n) {
long population = 0;
if ((n == 0) || (n == 1)) // base cases
{
return n;
} else // recursion step
{
population+=fib(n - 1) + fib(n - 2);
}
return population;
}
回答29:
Simple Fibonacci
public static void main(String[]args){
int i = 0;
int u = 1;
while(i<100){
System.out.println(i);
i = u+i;
System.out.println(u);
u = u+i;
}
}
}
回答30:
@chro is spot on, but s/he doesn't show the correct way to do this recursively. Here's the solution:
class Fib {
static int count;
public static void main(String[] args) {
log(fibWrong(20)); // 6765
log("Count: " + count); // 21891
count = 0;
log(fibRight(20)); // 6765
log("Count: " + count); // 19
}
static long fibRight(long n) {
return calcFib(n-2, 1, 1);
}
static long fibWrong(long n) {
count++;
if (n == 0 || n == 1) {
return n;
} else if (n < 0) {
log("Overflow!");
System.exit(1);
return n;
} else {
return fibWrong(n-1) + fibWrong(n-2);
}
}
static long calcFib(long nth, long prev, long next) {
count++;
if (nth-- == 0)
return next;
if (prev+next < 0) {
log("Overflow with " + (nth+1)
+ " combinations remaining");
System.exit(1);
}
return calcFib(nth, next, prev+next);
}
static void log(Object o) {
System.out.println(o);
}
}
来源：https://stackoverflow.com/questions/8965006/java-recursive-fibonacci-sequence