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: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
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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
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：
class Solution {
public int climbStairs(int n) {
int [] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
for(int i = 2; i <= n; i++){
dp[i] = dp[i-1] + dp[i-2];
}
return dp[n];
}
}