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
解题思路:
此题通过观察 n=2, 3, 4, ... 等步数的规律,发现是一个菲波那切数列,即 f(n) = f(n-1) + f(n-2),递归求解即可。
这里没有使用递归,采用动态规划的思想,只保留前两个状态。
Python实现:
# 动态规划
class Solution:
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
one = 0; two = 0; # 保留前两个状态
step = 1
while n > 0:
one = two
two = step
step = one + two
n -= 1
return step
a = 3
b = Solution()
print(b.climbStairs(5)) # 8 # f(5) = f(4) + f(3)
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