Climbing Stairs (LeetCode Dynamic Programming)

Question: 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?

Idea: Dynamic programming. For each stage i, the person can arrive from stage i-1 and i-2. So dp[i]=dp[i-1]+dp[i-2]. We can set dp[0]=1 to simplify the code, or set dp[1]=1, dp[2]=2.

Time: O(n) Space: O(n)

Code:

 public class Solution {  
   public int climbStairs(int n) {  
     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];  
   }  
 }