# 70. Climbing Stairs

Question link is here.

# Question

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

# Solution

A very easy dp question.

since $dp[i] = dp[i-1] + dp[i-2]$, Instead of use an array to store all value for 1 to n, we can just use two points to store $dp[i-1]$ and $dp[i-2]$. This help optimize space complexity.

class Solution {
    public int climbStairs(int n) {
        if(n <= 2)
            return n;
        int first = 1;
        int second = 2;
        int temp = 0;
        for(int i = 3; i <= n; i++) {
            temp = first + second;
            first = second;
            second = temp;
        }
        return temp;
    }
}

Time Complexity: $O(n)$.