> For the complete documentation index, see [llms.txt](https://garylai.gitbook.io/algorithm-and-data-structure/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://garylai.gitbook.io/algorithm-and-data-structure/problems/dynamic-programming/fibonacci-number.md).

# 509. Fibonacci Number

[509. Fibonacci Number](https://leetcode.com/problems/fibonacci-number/)

這一題是所有介紹動態規劃最基礎的一個題目，題目要求[斐波那契數](https://zh.wikipedia.org/wiki/%E6%96%90%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0) ，根據題目定義，可以很簡單的寫出遞迴的方法

## 遞迴

```python
class Solution:
    def fib(self, n: int) -> int:
        if n == 0:
            return 0
        elif n == 1:
            return 1
        else:
            return self.fib(n-1) + self.fib(n-2)
```

可是這個方法存在著很多的重複的子問題，所以可以用動態規劃的方法做。

## 自頂向下

```python
class Solution:
    def fib(self, n: int) -> int:
        table = defaultdict(int)
        table[1] = 1
        def traverse(n):
            # recursive
            if n <= 1:
                return table[n]
            if n in table:
                return table[n]
            else:
                table[n] = self.fib(n - 1) + self.fib(n - 2)
                return table[n]
        return traverse(n)
```

## 自底向上

```python
class Solution:
    def fib(self, n: int) -> int:
        memo = [0] * (n+1)
        memo[1] = 1
        for i in range(2, n+1):
            memo[i] = memo[i-1] + memo[i-2]

        return memo[n]
```

## 減少記憶體

```python
class Solution:    
    def fib(self, n: int) -> int:
        if n <= 1:
            return n
        first = 0
        second = 1
        for i in range(2, n + 1):
            first, second = second, first + second        
        return second
```


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