标签：style blog class code java c
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?
解析：设直至第n层有f(n)中走法，因为每次只能走1步或2步，则 f(n) = f(n-1) + f(n-2)
很明显是一个斐波那契数，f(0) = 0, f(1) = 1
1. 递归， f(n) = f(n-1) + f(n-2) 很容易想到递归，但是递归过程中存在大量的重复计算，会超时
2. 迭代
class Solution { public: int climbStairs(int n) { assert(n >= 0); int prevTwo = 0; int prevOne = 1; int cur = 1; for (int i = 1; i <= n; ++i) { prevOne = cur; cur = prevOne + prevTwo; prevTwo = prevOne; } return cur; } };
3. 直接计算：
斐波那契直接可以计算出表达数出来，且计算方法有好几种：等比数列法、矩阵法、生成函数法
具体见：斐波那契解法
class Solution { public: int climbStairs(int n) { assert(n >= 0); double s = sqrt(5); return floor((pow((1+s)/2, n+1) + pow((1-s)/2, n+1)) / s + 0.5); } };
Leetcode:Climbing Stairs 斐波那契数,布布扣,bubuko.com
Leetcode:Climbing Stairs 斐波那契数
标签：style blog class code java c
原文地址：http://www.cnblogs.com/wwwjieo0/p/3728745.html
