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文章目录
- 上楼梯问题：递归解法探究
- 问题定义
- 解决方案
- 1. 递归
- 2. 记忆化递归
- 关于python memo={}默认参数和字典的语法
- 语法功能
- 版本信息
- 注意事项
- 结论
上楼梯问题：递归解法探究
在这篇文章中，将对上楼梯问题进行深入探讨。上楼梯问题是一种常见的计算机科学问题，通常用于研究和理解动态规划、递归和记忆化等关键编程概念。首先，让我们更好地理解问题本身。
问题定义
给定一个楼梯，假设它有n个台阶。每次可以选择爬一阶、两阶或三阶。问题是，存在多少种可能的方式来爬到楼梯顶部？
这个问题在数学中被称为斐波那契变种，因为其解决方案包含了类似的递归结构。
解决方案
1. 递归
最直接的方法是使用递归。基本思想是每次爬楼梯都有三种选择，然后递归地求解剩余的台阶数。
def climbStairs(n):if n <= 0:return 0elif n == 1:return 1elif n == 2:return 2elif n == 3:return 4else:return climbStairs(n-1) + climbStairs(n-2) + climbStairs(n-3)
解释：因为每次爬楼梯都有三种选择，所以对于n阶楼梯，可以选择先走1阶，可以选择先走2阶，也可以选择先走3阶，那么求解爬n阶楼梯的方式数量就变成了：求解爬（n-1）阶楼梯的方式数量 + 爬（n-2）阶楼梯的方式数量 + 爬（n-3）阶楼梯的方式数量。
以上代码的缺点是运行时间会随着n的增大而呈指数增长，因为存在大量的重复计算。例如，climbStairs(n-1)和climbStairs(n-2)中都计算了climbStairs(n-3)。
2. 记忆化递归
为了避免重复计算，可以使用一种叫做"记忆化"的技术，即保存已经计算过的结果，当需要再次计算时，就直接从存储空间取出结果。
def climbStairs(n, memo={}):if n <= 0:return 0elif n == 1:return 1elif n == 2:return 2elif n == 3:return 4elif n not in memo:memo[n] = climbStairs(n-1) + climbStairs(n-2) + climbStairs(n-3)return memo[n]
此代码将以前计算的值存储在名为"memo"的字典中。如果需要计算的值已经在"memo"中，则直接返回该值，否则进行计算并将结果存入"memo"。
关于python memo={}默认参数和字典的语法
在Python中，memo={}是一种使用字典（dictionary）的方式来存储已经计算过的值，这样可以在以后需要这些值时快速获取，避免了重复计算。这种技术被称为"记忆化"（Memoization）。
语法功能
memo={}这部分代码创建了一个空的字典，名为memo。在函数climbStairs的参数列表中，将memo初始化为空字典。这样，当首次调用函数时，如果没有传入memo参数，那么就会创建一个新的空字典。
之后，在函数体中，当需要计算爬n个台阶的方法数时，先检查memo字典中是否已经有这个结果。如果有，就直接从字典中取出结果并返回；如果没有，就进行计算，并把结果存入memo字典，以备后续使用。
elif n not in memo:memo[n] = climbStairs(n-1) + climbStairs(n-2) + climbStairs(n-3)
return memo[n]
版本信息
这种默认参数和字典的使用方式在Python的早期版本中就已经存在。至少在Python 2.0中就已经支持，而这个版本于2000年发布。所以，可以说这是一个相当成熟的Python功能。
注意事项
需要注意的是，虽然这里memo作为默认参数被初始化为空字典，但由于Python默认参数的特性，即它们只会在函数定义的时候被初始化一次，所以在多次调用函数时，这个memo字典实际上是被共享的。这正好符合我们的需求，因为我们希望在多次递归调用中共享这个memo字典，以达到记忆化的目的。
结论
上楼梯问题提供了一个很好的框架，可以用来学习和理解递归和记忆化等编程技术。尽管这个问题看起来简单，但实际上包含了许多关键的计算机科学概念。
引用：
- “Programming Interviews Exposed: Secrets to Landing Your Next Job” by John Mongan et al.
- “Cracking the Coding Interview: 189 Programming Questions and Solutions” by Gayle Laakmann McDowell
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