# 70. Climbing Stairs

### 題目敘述

You are climbing a staircase. It takes `n` steps to reach the top.

Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?

Example 1:

``````Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
``````

Example 2:

``````Input: n = 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
``````

Constraints:

• `1 <= n <= 45`

Hint 1:

To reach nth step, what could have been your previous steps? (Think about the step sizes)

### 解法解析

#### 程式範例

##### Python
``````class Solution:
def climbStairs(self, n: int) -> int:
a, b = 1, 1
for _ in range(n):
a, b = b, a + b
return a
``````
##### JavaScript
``````/**
* @param {number} n
* @return {number}
*/
var climbStairs = function (n) {
if (n < 2) {
return n;
}
let first = 1;
let second = 2;
for (let i = 3; i <= n; i++) {
const third = first + second;
first = second;
second = third;
}
return second;
};
``````
##### Go
``````func climbStairs(n int) int {
if n <= 2 {
return n
}

n1, n2 := 1, 2

for i := 3; i <= n; i++ {
n2, n1 = n1+n2, n2
}
return n2
}
``````
##### Swift
``````class Solution {
func climbStairs(_ n: Int) -> Int {
if n < 3 { return n }

var cur = 2
var pre = 1
for i in 3...n {
let temp = pre
pre = cur
cur = cur + temp
}
return cur
}
}
``````
##### Kotlin
``````class Solution {
fun climbStairs(n: Int): Int {
return when {
n < 2 -> n
else -> {
var first = 1
var second = 2

for (i in 3..n) {
val temp = first + second
first = second
second = temp
}

second
}
}
}
}
``````