[Pyhon3] LeetCode 509. 斐波那契数列 - [Easy]

509. Fibonacci Number、509. 斐波那契数
（剑指offer - 10. 斐波那契数列）

题目：

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.

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。

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

Example 2:

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

Example 3:

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

Note:

0 ≤ N ≤ 30.

解答：

方法一：递归，时间复杂度为n的指数级，效率很低。不推荐。
方法二：循环，从下往上计算，将数列中间项保存起来。时间复杂度 O(n)。

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