动态规划之斐波那契数列

最近开始学习动态规划（dp），看了《算法导论》关于动态规划的内容，慢慢也整理一些dp的经典问题上来。
适合动态规划方法求解的最优化问题应该具备的两个要素：最优子结构和子问题重叠。dp的方法相当于是把要重复求解的子问题保存下来，这样就不用重复计算了，可以节省大量的时间。
这篇博客主要讲一下斐波那契数列的递归实现和动态规划实现。这个比较简单，直接上代码：
1）斐波那契数列的递归实现

def feibo_dg(n):
    if n == 1:
        return 1
    elif n == 2:
        return 2
    elif n > 2:
        return feibo_dg(n-1) + feibo_dg(n-2)

1）斐波那契数列的动态规划实现

def feibo_dp(n):
    f = []
    f.append(1)
    f.append(2)
    if n > 2:
        for i in range(2, n):
            f.append(f[i-1] + f[i-2])
    return f[n-1]

时间比较：

import time
if __name__ == '__main__':
    n = 41
    t1 = time.time()
    dg = feibo_dg(n)
    t2 = time.time()
    print ('dg_time:{:.8f}'.format(t2-t1))

    t3 = time.time()
    dp = feibo_dp(n)
    t4 = time.time()
    print('dp_time:{:.8f}'.format(t4 - t3))
    print ('dg result:', dg)
    print('dp result:', dp)

结果如下：

dg_time:62.57764745
dp_time:0.00000000
dg result: 267914296
dp result: 267914296

其实，经典的爬楼梯问题就是斐波那契数列问题。下面提供Leetcode上的一道原题和python解决方案：

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

我的python解决方案（其实和上面的斐波那契数列的动态规划解法一样）：

class Solution:
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n <= 0:
            return 0
        f = [0 for i in range(n)]
        if n >= 1:
            f[0] = 1
        if n >= 2:
            f[1] = 2
        if n > 2:
            for i in range(2, n):
                f[i] = f[i-1]+f[i-2]
        return f[n-1]

动态规划之斐波那契数列

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