【LeetCode】70. Climbing Stairs 解題報告

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70. Climbing Stairs / Easy

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

思路

爬梯子這題是很經典的 DP 題目，適合用來理解 DP 的概念。
每次只能爬 1 階或 2 階，意思就是如果要爬到第 3 階，只要能夠爬到第 1 階或第 2 階就可以了。
至於幾種爬法，只要將前兩階的爬法加起來就是現在這階的爬法數量。
也就是說，狀態轉移方程式為
dp[i] = dp[i-1] + dp[i-2]
而初始狀態則是第 1 與第 2 階，都是一種爬法。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    int climbStairs(int n) {
        vector<int> dp(n+1, 0);
        dp[0] = 1; dp[1] = 1;
        for(int i = 2; i <= n; ++i) {
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
};