Climbing Stairs

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?

 

仔细想想这其实就是一个斐波那契数列，这样一来问题就变成求斐波那契额数列第n个数的值。

一开始用递归发现效率低导致超时，于是用一个数组来存数列，代码如下：

public class Solution {
    public int climbStairs(int n) {
       ArrayList<Integer> list = new ArrayList<Integer>();
       list.add(1);
       list.add(2);
       if(n<=2){
           return list.get(n-1);
       }
       for(int i=2;i<n;i++){
           int one = list.get(i-2);
           int two = list.get(i-1);
           int sum = one+two;
           list.add(sum);
       }
       return list.get(n-1);
    }
}

 另外一种节省空间的方法：

public class Solution {
    public int climbStairs(int n) {
       if(n<=3){
           return n;
       }
       int preOne=1,preTwo=2,sum=0;
       for(int i=2;i<n;i++){
           sum = preOne+preTwo;
           preOne = preTwo;
           preTwo = sum;
       
       }
       return sum;
    }
}