70.climbing-stairs

Statement

简体中文English

Metadata

• Link: 爬楼梯
• Difficulty: Easy
• Tag: 记忆化搜索 数学 动态规划

假设你正在爬楼梯。需要 n 阶你才能到达楼顶。

每次你可以爬 1 或 2 个台阶。你有多少种不同的方法可以爬到楼顶呢？

 

示例 1：

输入：n = 2
输出：2
解释：有两种方法可以爬到楼顶。
1. 1 阶 + 1 阶
2. 2 阶

示例 2：

输入：n = 3
输出：3
解释：有三种方法可以爬到楼顶。
1. 1 阶 + 1 阶 + 1 阶
2. 1 阶 + 2 阶
3. 2 阶 + 1 阶

 

提示：

• 1 <= n <= 45

Metadata

• Link: Climbing Stairs
• Difficulty: Easy
• Tag: Memoization Math Dynamic Programming

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

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

 

Example 1:

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

Example 2:

Input: n = 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

 

Constraints:

• 1 <= n <= 45

Solution

Python 3

class Solution:
    f = {}

    def dfs(self, n: int) -> int:
        if n < 0:
            return 0
        if n == 0:
            return 1

        if n not in self.f.keys():
            self.f[n] = self.dfs(n - 1) + self.dfs(n - 2)

        return self.f[n]

    def climbStairs(self, n: int) -> int:
        return self.dfs(n)

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最后更新: October 11, 2023