coder-xiaomo/leetcode-problemset
1
0
Fork 0
mirror of https://gitee.com/coder-xiaomo/leetcode-problemset synced 2025-10-19 20:16:48 +08:00
Code Issues Projects Releases Wiki Activity GitHub Gitee

update

Browse Source
This commit is contained in:
小墨
2022-03-29 12:43:11 +08:00
parent 58bbdfd57c
commit 2b0511d272
10721 changed files with 8123 additions and 8119 deletions
Download Patch File Download Diff File Expand all files Collapse all files

42
leetcode-cn/problem (Chinese)/斐波那契数 [fibonacci-number].html Normal file
View File

@@ -0,0 +1,42 @@
<p><strong>斐波那契数</strong>&nbsp;（通常用&nbsp;<code>F(n)</code> 表示）形成的序列称为 <strong>斐波那契数列</strong> 。该数列由&nbsp;<code>0</code> 和 <code>1</code> 开始，后面的每一项数字都是前面两项数字的和。也就是：</p>
<pre>
F(0) = 0，F(1)&nbsp;= 1
F(n) = F(n - 1) + F(n - 2)，其中 n &gt; 1
</pre>
<p>给定&nbsp;<code>n</code> ，请计算 <code>F(n)</code> 。</p>
<p>&nbsp;</p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>n = 2
<strong>输出：</strong>1
<strong>解释：</strong>F(2) = F(1) + F(0) = 1 + 0 = 1
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>n = 3
<strong>输出：</strong>2
<strong>解释：</strong>F(3) = F(2) + F(1) = 1 + 1 = 2
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>n = 4
<strong>输出：</strong>3
<strong>解释：</strong>F(4) = F(3) + F(2) = 2 + 1 = 3
</pre>
<p>&nbsp;</p>
<p><strong>提示：</strong></p>
<ul>
<li><code>0 &lt;= n &lt;= 30</code></li>
</ul>