标签：blog one each ase 题目 cli note class ret
题目如下：
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.
思路：
当n＝1，结果为1
当n＝2，结果为2
当n＝3，结果为3
当n＝4，结果为7
易递推[n]=[n-1]+[n-2]，是一个斐波纳切数列。
本题代码：
int climbStairs(int n) { if (n<=0) return 0; if (n==1) return 1; if (n==2) return 2; int one_step_before = 2; int two_steps_before = 1; int all_ways = 0; for(int i=3; i<=n; i++) { all_ways = one_step_before + two_steps_before; two_steps_before = one_step_before; one_step_before = all_ways; } return all_ways; }
标签：blog one each ase 题目 cli note class ret
原文地址：http://www.cnblogs.com/pinganzi/p/6664135.html
