leetcode problem 70. Climbing Stairs-优快云博客

本文探讨了经典的爬楼梯问题，使用动态规划方法求解到达楼顶的不同路径数量。通过分析每次爬升1或2阶的情况，给出了递推公式，并提供了一个C++实现示例。该解决方案在在线评测中表现出色，运行速度和内存使用均优于100%的提交。

Description

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

Solution

consider the final step:

if the final step is one step, the ways to reach the top the ways[n-1]
if the final step is two step, the ways to reach the top the ways[n-2]
and there is no other way.

class Solution {
public:
    int climbStairs(int n) {
        int ways[1000]={0};
        ways[1] = 1; 
        ways[2] = 2;
        for(int i = 3; i <= n ; ++i)
        {
            ways[i] = ways[i-1] + ways[i-2];
        }
        return ways[n];
    }
};

result:

Runtime: 4 ms, faster than 100.00% of C++ online submissions for Climbing Stairs.
Memory Usage: 8.1 MB, less than 100.00% of C++ online submissions for Climbing Stairs.

20190410

tbc