剑指Offer面试题9 & Leetcode70

剑指Offer面试题9 & Leetcode70

Climbing Stairs  斐波那契数列

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?

解题思路

  考虑：爬楼梯的方法数相当于斐波那契数列求序号n处数列值。可以使用动态规划，写出递推关系式f(n)=f(n-1)+f(n-2)，用两个int值保存f(n-1)和f(n-2)，注意递推的初始值，即一阶台阶的方法数为1，二阶台阶的方法数为2，循环求出值。

Solution

public int climbStairs(int n) {
        if(n==1)    return 1;
        else if(n==2)   return 2;
        else{
            int num_n_1 = 2;
            int num_n_2 = 1;
            for(int i=3;i<=n;i++){
                int temp = num_n_1 + num_n_2;
                num_n_2 = num_n_1;
                num_n_1 = temp;
            }
            return num_n_1;
        }

    }