标签:.net 技术分享 bing http distinct bst cpp family sdn
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?
有f(n) = f(n - 1) + f(n - 2)。
这是一个斐波那契数列。
方法1。递归,太慢。
方法2。迭代。
方法3,数学公式。斐波那契数列的通项公式为
class Solution { public: int climbStairs(int n) { int prev = 0; int cur = 1; for(int i=1;i<=n;i++){ int tmp=cur; cur+=prev; prev=tmp; } return cur; } }; class Solution { public: int climbStairs(int n) { double s = sqrt(5); return floor((pow((1+s)/2, n+1) + pow((1-s)/2, n+1))/s + 0.5); } };
标签:.net 技术分享 bing http distinct bst cpp family sdn
原文地址:http://www.cnblogs.com/jzssuanfa/p/6973129.html