climbing stairs leetcode java

问题描述：

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?

提示：Dinamic Programming，动态规划，这是个斐波那契数列问题。

方法一：

//找规律，设n个台阶的方法数为f(n)
//f(1) = 1;f(2) = 2; .... ,从第n-1走到第n台阶，有两种方式：一步到或者两步到。
//因此，f(n) = f(n-1) + f(n-2); (n >= 3)
//递归运算，时间复杂度太高
public static int climbs(int n){
     if(n == 0)

        return 0;

     if(n == 1)
         return 1;
     if(n == 2)
         return 2;
     return climbs(n - 1) + climbs(n - 2); //分类
}

方法二：时间复杂度O(n)，空间复杂度O(1)

public int climbStairs(int n) {    

      if(n == 0)
           return 0;
      int prev = 0; //prev = f(n-2)
      int cur = 1; //cur = f(n-1)
      for(int i = 1; i <= n ; ++i){
           int tmp = cur;
           cur = prev + cur; // f(n) = f(n-2) + f(n-1)
           prev = tmp;
      }
      return cur;

}