[LeetCode] 70. Climbing Stairs_ Easy tag: Dynamic Programming

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

这个题目就是找到方程 f(i) = f(i - 1) + f(i - 2), i >= 2,然后用dynamic programming， 不过可以用滚动数组去优化。S：O(1)

Code

class Solution:
    def climbStairs(self, n):
        mem = list(range(1, n + 1)) # in python3 , range is an iterator
        for i in range(2, n):
            mem[i] = mem[i - 1] + mem[i - 2]
        return mem[n - 1]

 

滚动数组

class Solution:
    def climbStairs(self, n):
        mem = list(range(1, 4)) # in python3 , range is an iterator
        for i in range(2, n):
            mem[i%3] = mem[i%3 - 1] + mem[i%3 - 2]
        return mem[(n - 1)%3]

 

滚动数组2

class Solution:
    def climbStairs(self, n):
        dp = [1, 2]
        for i in range(2, n):
            dp[i %2] = dp[(i - 1)%2] + dp[(i - 2)%2]
        return dp[(n - 1)%2]