Climbing Stairs [LEETCODE]

最新推荐文章于 2020-12-11 14:59:21 发布

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最新推荐文章于 2020-12-11 14:59:21 发布

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原文链接：http://www.cnblogs.com/scenix/p/3368735.html

本文探讨了爬楼梯问题的动态规划方法，通过逐步分析和优化，从递归到迭代，实现高效求解路径数量。重点介绍了如何利用动态规划减少时间复杂度，避免重复计算，最终提供了一个更高效的解决方案。

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?

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Fisrt, I tried recursive way.

1 class Solution {
2 public:
3     int climbStairs(int n) {
4         // Recursively
5         if(0 == n) return 0;
6         if(1 == n) return 1;
7         return climbStairs(n - 1) + climbStairs(n - 2);
8     }
9 };

But this solution exceed time limit.

So let's think about it. One step has only 1 way , 2 steps have two ways.

Thus N steps have two ways to reach destination: N-1 steps plus one step; N-2 steps plus two steps.

So S[n] = S[n-1] + S[n-2]

Here' s the code:

 1 class Solution {
 2 public:
 3     int climbStairs(int n) {
 4         // Recursively
 5         if(0 == n) return 0;
 6         if(1 == n) return 1;
 7         int *step = new int[n+1];
 8         step[0] = 0;
 9         step[1] = 1;
10         step[2] = 2;
11         for(int i =3; i <=n ; i++){
12             step[i] = step[i-1] + step[i-2];
13         }
14         return step[n];
15     }
16 };