75 lines
2.3 KiB
Java
75 lines
2.3 KiB
Java
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# 10.1 斐波那契数列
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[NowCoder](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)
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## 题目描述
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求斐波那契数列的第 n 项，n <= 39。
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<!--<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> -->
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<div align="center"> <img src="pics/45be9587-6069-4ab7-b9ac-840db1a53744.jpg" width="300px"> </div><br>
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## 解题思路
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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。
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<div align="center"> <img src="pics/c13e2a3d-b01c-4a08-a69b-db2c4e821e09.png" width="350px"/> </div><br>
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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。
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```java
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public int Fibonacci(int n) {
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if (n <= 1)
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return n;
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int[] fib = new int[n + 1];
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fib[1] = 1;
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for (int i = 2; i <= n; i++)
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fib[i] = fib[i - 1] + fib[i - 2];
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return fib[n];
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}
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```
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考虑到第 i 项只与第 i-1 和第 i-2 项有关，因此只需要存储前两项的值就能求解第 i 项，从而将空间复杂度由 O(N) 降低为 O(1)。
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```java
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public int Fibonacci(int n) {
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if (n <= 1)
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return n;
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int pre2 = 0, pre1 = 1;
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int fib = 0;
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for (int i = 2; i <= n; i++) {
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fib = pre2 + pre1;
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pre2 = pre1;
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pre1 = fib;
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}
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return fib;
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}
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```
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由于待求解的 n 小于 40，因此可以将前 40 项的结果先进行计算，之后就能以 O(1) 时间复杂度得到第 n 项的值。
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```java
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public class Solution {
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private int[] fib = new int[40];
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public Solution() {
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fib[1] = 1;
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for (int i = 2; i < fib.length; i++)
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fib[i] = fib[i - 1] + fib[i - 2];
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}
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public int Fibonacci(int n) {
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return fib[n];
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}
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}
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```
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-1.png"></img></div>
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