[LeetCode July Challange]Day31-Climbing Stairs

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?

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

Constraints:

• 1 <= n <= 45

---

solution

time complexity : O(n)
space complexity : O(n)

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

動態歸劃，初始狀態樓梯為0階及1階時的走法皆為1次。(0階：不走)

n階樓梯的走法 = 走了1步後，剩下n-1階樓梯的走法 + 走了2步後，剩下n-2階樓梯的走法。

空間複雜度經設計後可達O(1)，但可讀性較差。