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动态规划（基础版）-- 斐波那契类型 -- 斐波那契数

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dynamic-programming/src/leetcode/editor/cn/FibonacciNumber.java Normal file
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//<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>
//
//<div><div>Related Topics</div><div><li>递归</li><li>记忆化搜索</li><li>数学</li><li>动态规划</li></div></div><br><div><li>👍 697</li><li>👎 0</li></div>
package leetcode.editor.cn;
// 509:斐波那契数
public class FibonacciNumber {
public static void main(String[] args) {
Solution solution = new FibonacciNumber().new Solution();
// TO TEST
}
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public int fib(int n) {
return n < 2 ? n : fib(n - 1) + fib(n - 2);
}
}
//leetcode submit region end(Prohibit modification and deletion)
}

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dynamic-programming/src/leetcode/editor/cn/doc/content/FibonacciNumber.md Normal file
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<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>
<div><div>Related Topics</div><div><li>递归</li><li>记忆化搜索</li><li>数学</li><li>动态规划</li></div></div><br><div><li>👍 697</li><li>👎 0</li></div>