509.fibonacci-number

Statement

• Difficulty: Easy
• Tag: `递归` `记忆化搜索` `数学` `动态规划`

``````F(0) = 0，F(1) = 1
F(n) = F(n - 1) + F(n - 2)，其中 n > 1
``````

``````输入：n = 2

``````

``````输入：n = 3

``````

``````输入：n = 4

``````

• `0 <= n <= 30`

• 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]
``````