https://github.com/yuanyu90221/leetcode_climbing_stairs

https://github.com/yuanyu90221/leetcode_climbing_stairs

Last synced: 16 days ago
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• Host: GitHub
• URL: https://github.com/yuanyu90221/leetcode_climbing_stairs
• Owner: yuanyu90221
• Created: 2021-07-16T15:56:28.000Z (over 3 years ago)
• Default Branch: master
• Last Pushed: 2021-07-16T15:57:23.000Z (over 3 years ago)
• Last Synced: 2024-06-20T22:14:21.701Z (7 months ago)
• Language: Go
• Size: 1.95 KB
• Stars: 0
• Watchers: 3
• Forks: 0
• Open Issues: 0
• Metadata Files:
  • Readme: README.md

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README

        
# leetcode_climbing_stairs

This repository is for solve [climbing-stairs](https://leetcode.com/problems/climbing-stairs/)

## problem description

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?

## Examples

### Example1

```

Input: n = 2

Output: 2

Explanation: There are two ways to climb to the top.

1. 1 step + 1 step

2. 2 steps

```

### Example2

```

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

```

## Observation

```

Notice that

There exists a relation

climb n stairs methods = first Step is One Step + first Step is Two Steps

                       = climb n - 1 stairs method + climb n - 2 stairs method

for n >= 3

                        

```

## implemtations

There are two implemention for solve this problem

Solution 1:

Create an array to store previous to accumlate which take Space Complexity O(n)

and iterate n step to compute n times

Thus, Time Complexity is O(n)

Solution 2:

Use iteration to accumlate previous two steps result

iterate n steps to compute 

Time Complexity is O(n)