# 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;
}
};