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?
思路:就是一个斐波那契数列,用动态规划解决
class Solution {
public:
int climbStairs(int n) {
int *F = new int[n + 1];
F[0] = 1;
F[1] = 1;
for(int i = 2; i <= n; ++i){
F[i] = F[i - 1] + F[i - 2];
}
return F[n];
}
};
本文探讨了一个经典的编程面试题——爬楼梯问题。通过使用动态规划的方法求解,文章详细解释了如何利用斐波那契数列来找出达到楼梯顶部的不同方式的数量。
218

被折叠的 条评论
为什么被折叠?



