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notes/10.1 斐波那契数列.md
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## 题目链接
[牛客网](https://www.nowcoder.com/practice/c6c7742f5ba7442aada113136ddea0c3?tpId=13&tqId=11160&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
[牛客网](https://www.nowcoder.com/practice/c6c7742f5ba7442aada113136ddea0c3?tpId=13\&tqId=11160\&tPage=1\&rp=1\&ru=/ta/coding-interviews\&qru=/ta/coding-interviews/question-ranking\&from=cyc\_github)
## 题目描述
求斐波那契数列的第 n 项，n \<= 39。
求斐波那契数列的第 n 项，n <= 39。
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?f(n)=\left\{\begin{array}{rcl}0&&{n=0}\\1&&{n=1}\\f(n-1)+f(n-2)&&{n>1}\end{array}\right." class="mathjax-pic"/></div> <br> -->
![](https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/45be9587-6069-4ab7-b9ac-840db1a53744.jpg)\
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/45be9587-6069-4ab7-b9ac-840db1a53744.jpg" width="330px"> </div><br>
## 解题思路
如果使用递归求解，会重复计算一些子问题。例如，计算 f(4) 需要计算 f(3) 和 f(2)，计算 f(3) 需要计算 f(2) 和 f(1)，可以看到 f(2) 被重复计算了。
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/c13e2a3d-b01c-4a08-a69b-db2c4e821e09.png" width="350px"/> </div><br>
![](https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/c13e2a3d-b01c-4a08-a69b-db2c4e821e09.png)\
递归是将一个问题划分成多个子问题求解，动态规划也是如此，但是动态规划会把子问题的解缓存起来，从而避免重复求解子问题。