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.
Example 1:
Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Example 2:
Input: 3 Output: 3 Explanation: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step
DP经典入门题,但是没想到第一次还超时了,因为直接递归的话肯定超时;
优化1: 用数组来存每一个步骤,减少计算量
优化2: 经过发现n只需要n-1和n-2这两个状态嘛,而且这三个值是连续的,也就是说数组的空间可以再优化成三个变量,当然看起来要费尽一点
class Solution { public: int climbStairs(int n) { if(n<3)return n; int step1=1,step2=2,res=0; for(int i=2;i<n;++i) { res=step1+step2; step1=step2; step2=res; } return res; } };