---
title: 'LeetCode 509. Fibonacci Number'
disqus: hackmd
---

# LeetCode 509. Fibonacci Number

## Description

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

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

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

## Constraints

0 <= n <= 30

## Answer

此題可用loop或是recurse來求解，在loop方面能用暫存的方式存現值和前一值來實現F(n) = F(n-1) + F(n-2)，在下一次的時候做疊加即可得答案。而recurse就是直接用公是解即可。

// 2021_12_01
// Loop solution
int fib(int N){
    if(N < 2){return N;}
    int ans = 0, cur = 1, pre = 0;
    for(int i = 2; i <= N; i++){
        ans = cur + pre;
        pre = cur;
        cur = ans;
    }
    return ans;
}

// recurse solution
int fib(int N){
    if(N == 0){return 0;}
    else if (N == 1){return 1;}
    else{return fib(N-1) + fib(N-2);}
}

## Link
https://leetcode.com/problems/fibonacci-number/
###### tags: `Leetcode`