70. Climbing stairs (simple)

Ideas:

The question asked how many different ways to climb to the top of the building , it is natural to think of using the Fibonacci sequence

Relying on the result of the previous time every time, think of using dp

 

Code:

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];
    }
}

 

break down:

1) To prevent subscript overflow, when dp is declared, the length is added 1 (n+1) on the basis of n

 

2) This is the first form of dp: ( linear )

Rely only on a limited number (two) of previous states