[leetcode] 70.Climbing Stairs
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题目:
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?
题意:
你可以往上爬一个楼梯或者两格楼梯,到达顶端一共有n格台阶,问你到达顶端一共有多少种可能。
思路:
这个题目比较简单,就是斐波拉契的变形,对于第一格台阶上去的可能有1种,对于第二格台阶上去的可能一共有两种,直接跨两格上去,或者从台阶一上跨一格上去。采用动态规划的方法,我们记第k格台阶有f(k)种方法到达,那么对于第n格台阶,一共有f(n) = f(n-1) + f(n-2)种方法可以到达(即从第n-1格台阶跨一步到达第n格台阶,或者从第n-2格台阶跨两步到达地n格台阶)。
以上。
代码如下:
class Solution {public: int climbStairs(int n) { if(n == 1)return 1; else if(n == 2)return 2; int a = 1, b = 2; for(int k = 3; k <= n; k++){ int temp = a; a = b; b += temp; } return b; }};
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