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【leetcode】509. Fibonacci Number

mon 提高效率 put alc int def return ret 效率

題目如下:

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.

解題思路:遞歸的經典例子,當然為了提高效率,可以緩存中間結果。這題好像以前是Premium的,最近解鎖了。

代碼如下:

class Solution(object):
    def fib(self, N):
        """
        :type N: int
        :rtype: int
        
""" if N == 0 or N == 1: return N return self.fib(N-1) + self.fib(N-2)

【leetcode】509. Fibonacci Number