# Leetcode 70. Climbing stairs

^{9}/Jun 2018

Let’s condsider easy dynamic problem: climbing stairs. This is Leetcode 70. Climbing stairs.

## Description

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. 1 step + 1 step
2. 2 steps
```

### Example 2

```
Input: 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
```

## Analysis

Number of distinct ways for each stair is a sum of number of distinct ways for previous stair and
number of distinct ways for previous-previous stair, recursively it can be expressed as following:

`ways(n)=ways(n-1) + ways(n-2)`

## Recursive solution

```
class Solution {
public int climbStairs(int n) {
switch(n) {
case 0: return 0;
case 1: return 1;
case 2: return 2;
default: return climbStairs(n-1) + climbStairs(n-2);
}
}
}
```

This recursive solution will do many repeating computations, but this is most clear and concise way to express solution. Now let’s consider more economical approach.

## Better way

This is better solution, with time complexity `O(n)`

and space complexity `O(1)`

:

```
class Solution {
public int climbStairs(int n) {
if (n <= 2) return n;
int prevPrevStair = 1, prevStair = 2;
int curStair=0;
for (int i=2; i<n; i++) {
curStair = prevStair + prevPrevStair;
prevPrevStair = prevStair;
prevStair = curStair;
}
return curStair;
}
}
```