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
这是一个简单的爬楼梯的题目,可以一次爬一层,也可以爬两层。如果爬一层就是i-1的情况,如果爬两层就是i-2的情况,所以
l[i]=l[i-1]+l[i-2]
class Solution:
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
l = [1 for i in range(n+1)]
for i in range(2,n+1):
l[i]=l[i-1]+l[i-2]
return l[n]
开始的时候我的写法是
class Solution:
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
if n==2:
return 2
if n==3:
return 3
return self.climbStairs(n-1)+self.climbStairs(n-2)
这样字节测试时通过的,但是放到leetcode里面显示递归超过了最大深度,应该是递归的深度太深,而且大量节点重复计算。