迭代法爬楼梯python
时间: 2023-10-12 07:08:01 浏览: 141
迭代法爬楼梯的Python实现如下所示:
```python
class Solution:
def climbStairs(self, n: int) -> int:
dp = [0 for _ in range(n+1)]
dp = 1
dp = 1
for i in range(2, n+1):
dp[i] = dp[i-1] + dp[i-2]
return dp[n]
```
这段代码使用动态规划的思想来解决爬楼梯的问题。dp[i]表示爬到第i级台阶的方法数。初始时,dp和dp都为1,表示爬到第0级和第1级台阶只有一种方法。
通过迭代计算,dp[i]可以由dp[i-1]和dp[i-2]得到,因为每一步只能走一个台阶或者两个台阶。最终返回dp[n]即可得到爬到第n级台阶的方法数。<span class="em">1</span><span class="em">2</span><span class="em">3</span>
#### 引用[.reference_title]
- *1* *3* [python 爬楼梯 动态规划 算法 leetcode](https://blog.csdn.net/qq_50974223/article/details/112889642)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 50%"]
- *2* [Leetcode 爬楼梯 (Python解法)](https://blog.csdn.net/weixin_42095500/article/details/82951947)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 50%"]
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