15算法课程 70. Climbing Stairs
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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.
solution:
这个题目是一个计算n层阶梯情况下,走到顶端的路径种数(要求每次只能上1层或者2层阶梯)。
这是一个动态规划的题目:
n = 1 时 ways = 1;
n = 2 时 ways = 2;
n = 3 时 ways = 3;
…
n = k 时 ways = ways[k-1] + ways[k-2];
明显的,这是著名的斐波那契数列问题。有递归和非递归两种方式求解,但是亲测递归方式会出现 The Time Limit异常,所以只能采用非递归计算,可以用一个动态数组保存结果。
code:
class Solution {public: int climbStairs(int n) { if (n <= 0) return 0; else if (n == 1) return 1; else if (n == 2) return 2; int *r = new int[n]; r[0] = 1; r[1] = 2; for (int i = 2; i < n; i++) r[i] = r[i - 1] + r[i - 2]; int ret = r[n - 1]; delete []r; return ret; }};
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