# 509.Fibonacci-Number

## ้ข็ฎๅฐๅ

https://leetcode.com/problems/fibonacci-number/

## ้ข็ฎๆ่ฟฐ

``````The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

F(0) = 0,   F(1) = 1
F(N) = F(N - 1) + F(N - 2), for N > 1.
Given N, calculate F(N).

Example 1:
Input: 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 2:
Input: 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.

Example 3:
Input: 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.

Note:
0 โค N โค 30.``````

## ไปฃ็ 

### Approach #1 Recursion

Complexity Analysis

• Time complexity : O(2^N)

• Space complexity : O(N)

``````class Solution {
public int fib(int N) {
if (N <= 1) {
return N;
}
return fib(N - 1) + fib(N - 2);
}
}``````

### Approach #2 Bottom-Up Approach using memoization

``````class Solution {
public int fib(int N) {
if (N <= 1) {
return N;
}
return memoize(N);
}

public int memoize(int N) {
int[] cache = new int[N + 1];
cache[1] = 1;

for (int i = 2; i <= N; i++) {
cache[i] = cache[i - 1] + cache[i - 2];
}
return cache[N];
}
}``````

### Approach #3 Top-Down Approach using Memoization

``````class Solution {
private Integer[] cache = new Integer[31];

public int fib(int N) {
if (N <= 1) {
return N;
}
cache[0] = 0;
cache[1] = 1;
return memoize(N);
}

public int memoize(int N) {
if (cache[N] != null) {
return cache[N];
}
cache[N] = memoize(N-1) + memoize(N-2);
return memoize(N);
}
}``````

### Approach #4 Iterative Top-Down Approach

``````class Solution {
if (N <= 1) {
return N;
}
if (N == 2) {
return 1;
}

int current = 0;
int prev1 = 1;
int prev2 = 1;

for (int i = 3; i <= N; i++) {
current = prev1 + prev2;
prev2 = prev1;
prev1 = current;
}

return current;
}``````

### Approach #5 Matrix Exponetiation

``````class Solution {
int fib(int N) {
if (N <= 1) {
return N;
}
int[][] A = new int[][]{{1, 1}, {1, 0}};
matrixPower(A, N-1);

return A[0][0];
}

void matrixPower(int[][] A, int N) {
if (N <= 1) {
return;
}
matrixPower(A, N/2);
multiply(A, A);

int[][] B = new int[][]{{1, 1}, {1, 0}};
if (N%2 != 0) {
multiply(A, B);
}
}

void multiply(int[][] A, int[][] B) {
int x = A[0][0] * B[0][0] + A[0][1] * B[1][0];
int y = A[0][0] * B[0][1] + A[0][1] * B[1][1];
int z = A[1][0] * B[0][0] + A[1][1] * B[1][0];
int w = A[1][0] * B[0][1] + A[1][1] * B[1][1];

A[0][0] = x;
A[0][1] = y;
A[1][0] = z;
A[1][1] = w;
}
}``````

### Approach #6 Math

``````class Solution {
public int fib(int N) {
double goldenRatio = (1 + Math.sqrt(5)) / 2;
return (int)Math.round(Math.pow(goldenRatio, N)/ Math.sqrt(5));
}
}``````

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