# 70. Climbing Stairs (Easy)

https://leetcode.com/problems/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
```

## Solutions

### 1.

``````class Solution {
public int climbStairs(int n) {
if (n == 0 || n == 1) {
return 1;
}

if (n == 2) {
return 2;
}

// there are n stairs, actually, the walking path start from 0, the n
int[] pathWay = new int[n];

// stand still, 1 possibility
pathWay[n - 1] = 1;

// from n-2 to n, 2 possibility
pathWay[n - 2] = 2;

for (int i = n - 3; i >= 0; i--) {
pathWay[i] = pathWay[i + 1] + pathWay[i + 2];
}

return pathWay[0];
}
}
``````

### 2.

``````class Solution {
// pay attention that there is n steps, suppose the topmost stair numbered 0,
// which means your starting stair's number is n, rather than n-1
int[] pathWay;

public int climbStairs(int n) {
if (n == 0 || n == 1) {
return 1;
}

if (n == 2) {
return 2;
}

pathWay = new int[n + 1];
Arrays.fill(pathWay, -1);

pathWay[1] = 1;
pathWay[2] = 2;

// set off from n to 0
return walkThrough(n);
}

public int walkThrough(int n) {
if (n <= 0) {
return 0;
}

if (pathWay[n] >= 0) {
return pathWay[n];
}

int p1 = walkThrough(n - 1);
int p2 = walkThrough(n - 2);

pathWay[n] = p1 + p2;

return p1 + p2;
}
}
``````