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? Hide Tags: Dynamic Programming Solution:一个台阶的方法次数为1次,两个台阶的方法次数为2个.n个台阶的方法可以理解成上n-2…
一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 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…
题目: 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. 题解: 爬到第n层的方法要么是从第n-1层一步上来的,要不就是从n-2层2步…
Leetcode 70 Climbing Stairs Easy https://leetcode.com/problems/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? Note…
#Method1:动态规划##当有n个台阶时,可供选择的走法可以分两类:###1,先跨一阶再跨完剩下n-1阶:###2,先跨2阶再跨完剩下n-2阶.###所以n阶的不同走法的数目是n-1阶和n-2阶的走法数的和class Solution(object): def climbStairs(self, n): if n==1 or n==2 or n==0: return n steps=[1,1] for i in xrang…
一共有n个台阶,每次跳一个或者两个,有多少种走法,典型的Fibonacii问题 class Solution(object): def climbStairs(self, n): if n<0: return 0 if n<2: return 1 first,second = 1,1 for v in range(2,n+1): res = first+second first,second = return res 还有一种,每次可以跳任意阶,有2^(n-1)种跳法…
