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509.fibonacci-number

Statement

Metadata
  • Link: 斐波那契数
  • Difficulty: Easy
  • Tag: 递归 记忆化搜索 数学 动态规划

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 01 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

Metadata
  • Link: Fibonacci Number
  • Difficulty: Easy
  • Tag: Recursion Memoization Math Dynamic Programming

The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

F(0) = 0, F(1) = 1
F(n) = F(n - 1) + F(n - 2), for n > 1.

Given n, calculate F(n).

 

Example 1:

Input: n = 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 2:

Input: n = 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.

Example 3:

Input: n = 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.

 

Constraints:

  • 0 <= n <= 30

Solution

class Solution:
    def fib(self, n: int) -> int:
        f = [0 for i in range(n + 2)]
        f[1] = 1
        for i in range(2, n + 1):
            f[i] = f[i - 1] + f[i - 2]
        return f[n]

最后更新: January 15, 2023
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