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.

Input: 2Output: 2Explanation: There are two ways to climb to the top.1. 1 step + 1 step2. 2 steps -----------------------------------------------------------

Input: 3Output: 3Explanation: There are three ways to climb to the top.1. 1 step + 1 step + 1 step2. 1 step + 2 steps3. 2 steps + 1 step

 

题意：

有n阶台阶，可以每次爬1步或者2步， 爬到终点有多少种不同的方法

 

思路：

一维dp

用一个数组存下当前 k (k<=n)阶台阶有多少种不同的走法

发现规律

爬上n阶台阶 = 爬上n-1阶台阶再爬1阶 + 爬上n-2阶台阶再爬2阶

爬上n-1阶台阶 =  爬上n-2阶台阶再爬1阶  +  爬上n-3阶台阶再爬2阶

爬上n-2阶台阶 =  爬上n-3阶台阶再爬1阶  +  爬上n-4阶台阶再爬2阶

初始化，

dp[0] = 1

dp[1] = 1

转移方程，

dp[i] = dp[i-1] + dp[i-2] 

 

代码：

1 class Solution { 2     public int climbStairs(int n) { 3         int [] dp = new int[n + 1]; 4         dp[0] = 1; 5         dp[1] = 1; 6         for(int i = 2; i <= n; i++){ 7             dp[i] = dp[i-1] + dp[i-2]; 8         } 9         10         return dp[n];   11     }12 }