# 70. Climbing Stairs

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

# Solution

Approach 1: Recursion.

Approach 2: Dynamic programming.

# Code (Python)

Approach 1:

Approach 2:

# Code (C++)

Approach 1:

// Recursion: Time Limit Exceeded.
class Solution {
public:
    int climbStairs(int n) {
        if (n == 0) return 0;
        if (n == 1) return 1;
        if (n == 2) return 2;
        return climbStairs(n-1) + climbStairs(n-2);
    }
};

Approach 2:

class Solution {
public:
    int climbStairs(int n) {
        if (n == 0) return 0;
        if (n == 1) return 1;
        if (n == 2) return 2;
        int dp[n+1];
        dp[0] = 1;
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; ++i)
        {
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
};

// Fibonacci sequence (with different base).
class Solution {
public:
    int climbStairs(int n) {
        if (n == 0) return 0;
        if (n == 1) return 1;
        if (n == 2) return 2;
        int p1 = 1;
        int p2 = 2;
        int res = 0;
        for (int i = 3; i <= n; ++i)
        {
            res = p1 + p2;
            p1 = p2;
            p2 = res;
        }
        return res;
    }
};