70. Climbing Stairs
[Problem]
You are climbing a staircase. It takes n steps to reach the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Example 1:
Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: n = 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
Constraints:
- 1 <= n <= 45
[Solution]
class Solution:
def climbStairs(self, n: int) -> int:
dp = [] # dp[n] = dp[n-1] + dp[n-2]
dp.insert(0, 0), dp.insert(1, 1), dp.insert(2, 2)
for i in range(3, n+1):
dp.insert(i, dp[i-1] + dp[i-2])
return dp[n]
dp[n] is ways to climb to the top.
Suppose that you stand on "n steps" and go down to "0 steps". Now we will choice 1 steps or 2 steps to get to "0 steps" staircase.
| 0 | 1 | 2 | 3 | ... | n-2 | n-1 | n | ||
If you choice 1 steps to go down, you will reach "n-1 steps" of staircase, and If you choice 2 steps to go down, you will reach "n-2 steps" of staircase.
It means that ways of reaching "n steps" are equal to the sum of dp[n-1] and dp[n-2].