# Dynamic Programming - 70. Climbing Stairs

70. Climbing Stairs

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

1. 确定状态，找出问题中的子问题，考虑最后一步是什么。
2. 写出状态转移方程。
3. 定义初始条件和边界条件（Corner Case）
4. 计算顺序，也就是从小到大还是从大到小，爬楼梯问题很明显，是从小到大，备忘录类型。
这题中，我的dp[i]代表的是到达第i-1台台阶需要走几步，所以dp[0]代表的是第一台台阶需要的步数，所以我的slice长度开辟了n。

go：

``````func climbStairs(n int) int {

if n <= 0 {
return 0
}

if (n == 1 || n == 2) {
return n
}
dp := make([]int, n)
// initialization
dp[0] = 1
dp[1] = 2

for i := 2; i < n; i++ {
dp[i] = dp[i-1] + dp[i-2]
}

return dp[n-1]
}

``````

Nothing just happens, it's all part of a plan.