## Algorithm

Problem Name: 70. Climbing Stairs

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
```

Constraints:

• `1 <= n <= 45`

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
int climbStairs(int n) {
int p, pp, k, i;

if (n == 1) return 1;
if (n == 2) return 2;

p = 2;
pp = 1;

for (i = 3; i  < = n; i++) {
k = p + pp;
pp = p;
p = k;
}

return p;
}
``````
Copy The Code &

Input

cmd
n = 2

Output

cmd
2

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
int climbStairs(int n) {
vector<int>dp(n + 1);
dp[0] = 1;
dp[1] = 1;
for(int i = 2; i  <  n + 1; i++)
dp[i] = dp[i - 1] + dp[i - 2];
return dp[n];
}
};
``````
Copy The Code &

Input

cmd
n = 3

Output

cmd
3

### #3 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int climbStairs(int n) {
if (n == 1) {
return 1;
}
int[] dp = new int[n + 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];
}
}
``````
Copy The Code &

Input

cmd
n = 2

Output

cmd
2

### #4 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const climbStairs = function(n) {
const hash = {};
return single(n, hash);
};

function single(i, hash) {
if (hash.hasOwnProperty(i)) {
return hash[i];
}
if (i === 1) {
hash[1] = 1;
return 1;
}
if (i === 2) {
hash[2] = 2;
return 2;
}
hash[i] = single(i - 1, hash) + single(i - 2, hash);
return hash[i];
}
``````
Copy The Code &

Input

cmd
n = 2

Output

cmd
2

### #5 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def climbStairs(self, n: int) -> int:
memo = {}
def dfs(i):
if i >= n: return 1 if i == n else 0
if i not in memo:
memo[i] = dfs(i + 1) + dfs(i + 2)
return memo[i]
return dfs(0)
``````
Copy The Code &

Input

cmd
n = 3

Output

cmd
3

### #6 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _070_ClimbingStairs
{
public int ClimbStairs(int n)
{
if (n < 4) { return n; }

int x1 = 2, x2 = 3, temp;
for (int i = 4; i  < = n; i++)
{
temp = x1 + x2;
x1 = x2;
x2 = temp;
}

return x2;
}
}
}
``````
Copy The Code &

Input

cmd
n = 3

Output

cmd
3