70. Climbing Stairs ---E

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

A:
就是简单的DP问题

public class Solution {
        public int climbStairs(int n) {
           //DP
           if(n ==1 || n ==2)
            return n;
           int[] A = new int[n];
           A[0] = 1;
           A[1] = 2;
           for(int i=2;i<n;i++)
                A[i] = A[i-1]+A[i-2];
           return A[n-1];
        }
    }

Mistakes:
1: 哎，刚开始，竟然，连n=2的情况，都能弄错，   SB啊，真是~~~~~
应该是基于n=1 n=2的情况来考虑， 而不是因为，有了n=0 ,n =1 的情况了，就可以作为基本情况了。-----------------------------这个是base case 没有考虑清楚。


Learned: