Class Solution
Easy
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
To solve the “Climbing Stairs” problem in Java with the Solution class, follow these steps:
- Define a method 
climbStairsin theSolutionclass that takes an integernas input and returns the number of distinct ways to climb to the top of the staircase withnsteps. - Initialize an array 
dpof sizen+1to store the number of distinct ways to reach each step. - Set 
dp[0] = 1anddp[1] = 1since there is only one way to reach the first and second steps. - Iterate over the steps from 
2ton:- At each step 
i, the number of distinct ways to reach stepiis the sum of the number of ways to reach stepsi-1andi-2. - Store this sum in 
dp[i]. 
 - At each step 
 - Return 
dp[n], which represents the number of distinct ways to climb to the top of the staircase withnsteps. 
Here’s the implementation of the climbStairs method in Java:
class Solution {
    public int climbStairs(int n) {
        if (n == 1) return 1;
        
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        
        for (int i = 2; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        
        return dp[n];
    }
}
This implementation efficiently calculates the number of distinct ways to climb the stairs using dynamic programming, with a time complexity of O(n) and a space complexity of O(n).
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Constructor Summary
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Method Summary
 
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Constructor Details
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Solution
public Solution() 
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Method Details
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climbStairs
public int climbStairs(int n)  
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