[leetcode] 70. Climbing Stairs @ python

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于 2019-01-20 17:19:34 发布

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本文链接：https://blog.youkuaiyun.com/danspace1/article/details/86563813

版权

本文探讨了经典的爬楼梯问题，即如何计算达到第n个台阶的不同方式数量。通过使用Fibonacci数列的概念，我们发现这个问题可以转化为求解数列的第n项。为避免重复计算，采用字典存储已知结果，实现时间复杂度为O(n)，空间复杂度为O(1)的高效算法。

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原题

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 step + 1 step
2 steps
Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
3. 1 step + 1 step + 1 step
4. 1 step + 2 steps
5. 2 steps + 1 step

解法

fibonacci数列, 第n个台阶的爬行方法等于从第n-2个台阶开始爬2步上来 + 从第n-1个台阶开始爬1步上来.

f(n) = f(n-2) + f(n-1)

这里如果直接用递归会导致时间复杂度太大, 因此我们用字典保存已知的结果, 然后遍历算出结果.
Time: O(n)
Space: O(1)

代码

class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        known = {1: 1, 2:2}
        if n in known:
            return known[n]
        for i in range(3, n+1):
            known[i] = known[i-1] + known[i-2]
        return known[n]