---
tags: leetcode
---

# [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)
---
# My Solution
## Solution 1: Brute Force (TLE)
### The Key Idea for Solving This Coding Question
### C++ Code

class Solution {
public:
    int climbStairs(int n) {
        if (n <= 2) {
            return n;
        }
        
        return climbStairs(n - 1) + climbStairs(n - 2);
    }
};

### Time Complexity
$O(2^{n})$
### Space Complexity
$O(2^{n})$
## Solution 2: Top-down DP
### The Key Idea for Solving This Coding Question
### C++ Code

class Solution {
public:
    int climbStairs(int n) {
        vector<int> stepToCnt(n + 1, -1);
        
        stepToCnt[1] = 1;
        if (n == 1) {
            return stepToCnt[1];
        }
        
        stepToCnt[2] = 2;
        if (n == 2) {
            return stepToCnt[2];
        }
        
        return dp(n, stepToCnt);
    }

private:
    // dp returns the number of ways for climbing to the n-th step.
    int dp(int n, vector<int> &stepToCnt) {
        if (stepToCnt[n] > 0) {
            // We have calculated stepToCnt[n].
            // Just reuse its answer.
            return stepToCnt[n];
        }
        
        stepToCnt[n] = dp(n - 1, stepToCnt) + dp(n - 2, stepToCnt);
        return stepToCnt[n];
    }
};

### Time Complexity
$O(n)$
### Space Complexity
$O(2n) = O(n)$
## Solution 3: Buttom-up DP
### The Key Idea for Solving This Coding Question
### C++ Code

class Solution {
public:
    int climbStairs(int n) {
        vector<int> dp(n + 1, -1);
        
        dp[1] = 1;
        if (n == 1) {
            return dp[1];
        }

        dp[2] = 2;
        if (n == 2) {
            return dp[2];
        }
        
        for (int i = 3; i <= n; ++i) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        
        return dp[n];
    }
};

### Time Complexity
$O(n)$
### Space Complexity
$O(n)$
## Solution 4: Buttom-up DP
### The Key Idea for Solving This Coding Question
### C++ Code

class Solution {
public:
    int climbStairs(int n) {
        int a = 1;
        if (n == 1) {
            return a;
        }

        int b = 2;
        if (n == 2) {
            return b;
        }
        
        int c;
        for (int i = 3; i <= n; ++i) {
            c = a + b;
            a = b;
            b = c;
        }
        
        return c;
    }
};

### Time Complexity
$O(n)$
### Space Complexity
$O(1)$
# Miscellaneous
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