leetcode 509. Fibonacci Number

題目:

 
 留言

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.

題意:

使用程式解費波納氣數列

解法:

按照題意使用非常簡單的recursive function 可以輕易地解出,但是效率會非常差

class Solution:
def fib(self, n: int) -> int:
if n==0:
return 0
elif n==1:
return 1
else:
return self.fib(n-1)+self.fib(n-2)

 

使用hash table來記錄所有以前計算過的數列可以節省非常多時間

class Solution:
def fib(self, N: int) -> int:
if N <= 1:
return N
return self.memoize(N)
def memoize(self, N):
hash_table = {0: 0, 1: 1}
for i in range(2, N+1):
hash_table[i] = hash_table[i-1] + hash_table[i-2]
return hash_table[N]

一個演算法從後前算,一個從前網後算,效率差了30倍之多,足以體現演算法之美阿

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I0o0I
3 years ago

有時候很好奇,這種只看效率,但是不比較實際跑出的細項,是不是真的有算正確?
版主回覆:(02/09/2022 09:24:27 PM)
可以看細項吧? 每一個測驗項目都放在leetcode裡面