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?
class Solution {
public:
int climbStairs(int n) {
int *p = new int[n + 1];
p[0] = p[1] = 1;
for(int i = 2;i <= n;i ++)
p[i] = p[i - 1] + p[i - 2];
return p[n];
}
};
本文探讨了如何使用动态规划解决爬楼梯问题,通过创建一个动态数组来存储到达每个阶梯的不同方式,最终计算到达顶层的总方式数。此方法不仅高效地解决了问题,还展示了动态规划在解决递归问题时的实用性。
488

被折叠的 条评论
为什么被折叠?



