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
python实践
- 斐波那契数列
- 动态规划思想
#!/usr/bin/env python
# _*_ coding:utf-8 _*_
class Solution(object):
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
# 找规律，斐波那契数列
pre = 1
ppre = 1
if n == 1:
return 1
else:
for i in range(2, n + 1):
tmp = pre
pre = ppre +pre
ppre = tmp
return pre
def climbStairs2(self, n):
# 定义到第i个台阶,可能的情况数
if n == 1 or n == 2:
return n
else:
dp = [0]*(n+1)
dp[1] = 1
dp[2] = 2
for i in range(3, n+1):
dp[i] = dp[i-1] + dp[i-2]
return dp[-1]
s = Solution()
s2 = s.climbStairs2(5)
print(s2)
