70. Climbing Stairs

口起灵，比那个稍微难点，一次不是跳两级，是跳k级，给一个k，一个n，n是总台阶数，k是每次最多跳的个数，可以比它少。我一上来说dp，给了个时间复杂度 O(kn)，空间O(n)的。他说能不能简化一下空间。我说那我给了个时间kn, 空间k的。最后讨论了一阵子我简化成了时间n，空间min(n,k)的。就是不用每次都把那k个加一遍，保持一个sum就好了，每次删除一个旧的再加一个新的。




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?
Note: Given n will be a positive integer.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: 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




class Solution {
    public int climbStairs(int n) {
      if(n == 1) return 1; // corner case 
      if(n == 2) return 2;// corner case 
      int[] dp = new int[n + 1];
      dp[1] = 1;
      dp[2] = 2;
      for(int i = 3; i <= n; i++){
        dp[i] = dp[i -1] + dp[i - 2];
      }
      return dp[n];
    }
}