LeetCode 0509. Fibonacci Number斐波那契數【Easy】【Python】【動態規劃】
Problem
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: 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.
問題
斐波那契數,通常用 F(n) 表示,形成的序列稱爲斐波那契數列。該數列由 0 和 1 開始,後面的每一項數字都是前面兩項數字的和。也就是:
F(0) = 0, F(1) = 1
F(N) = F(N - 1) + F(N - 2), 其中 N > 1.
給定 N,計算 F(N)。
示例 1:
輸入:2
輸出:1
解釋:F(2) = F(1) + F(0) = 1 + 0 = 1.
示例 2:
輸入:3
輸出:2
解釋:F(3) = F(2) + F(1) = 1 + 1 = 2.
示例 3:
輸入:4
輸出:3
解釋:F(4) = F(3) + F(2) = 2 + 1 = 3.
提示:
0 ≤ N ≤ 30
思路
解法一:遞歸
fib(n) = fib(n - 1) + fib(n - 2)
注意,fib(n)會越界,所以最好是:
fib(n) % 1000000007 = (fib(n - 1) % 1000000007 + fib(n - 2) % 1000000007) % 1000000007
但是因爲 Python 中整形數字的大小限制取決計算機的內存(可理解爲無限大),因此可不考慮大數越界問題。
Python3代碼
class Solution:
def fib(self, n: int) -> int:
# solution one: 遞歸
if n == 0:
return 0
if n == 1:
return 1
return (self.fib(n - 1) + self.fib(n - 2)) % 1000000007
解法二:動態規劃
時間複雜度: O(n)
空間複雜度: O(1)
Python3代碼
class Solution:
def fib(self, n: int) -> int:
# solution two: 動態規劃
dp_0, dp_1 = 0, 1
for _ in range(n):
dp_0, dp_1 = dp_1, dp_0 + dp_1
return dp_0 % 1000000007