carloscn / carloscn/structstudy

leetcode509:斐波那契数(fibonacci-number)

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duplicate Lang-c/c++ Level-easy 动态规划 数组 查找
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Description

### 问题描述

斐波那契数 (通常用 F(n) 表示)形成的序列称为 斐波那契数列 。该数列由 0 和 1 开始,后面的每一项数字都是前面两项数字的和。也就是:

F(0) = 0,F(1) = 1
F(n) = F(n - 1) + F(n - 2),其中 n > 1
给定 n ,请计算 F(n) 。

示例 1:

输入:n = 2
输出:1
解释:F(2) = F(1) + F(0) = 1 + 0 = 1
示例 2:

输入:n = 3
输出:2
解释:F(3) = F(2) + F(1) = 1 + 1 = 2
示例 3:

输入:n = 4
输出:3
解释:F(4) = F(3) + F(2) = 2 + 1 = 3
 
提示:

0 <= n <= 30

来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/fibonacci-number

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