LeetCode 70. Climbing Stairs

原題連接在這裏：https://leetcode.com/problems/climbing-stairs/

題目：

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.

題解：

實際上是Fibonacci Number. e.g. n = 100, 假設登到99臺階有m種方法，登到98臺階有n種方法，那麼從99到100都是上一步，因此仍是m種方法. 從98臺階登到100都是一次登兩個臺階，仍是n種方法，若果在98登一個臺階，就是到了99, 這種方法已經包含在最初登到99臺階的m種方法中了。

因此登到100的方法就是m+n.

Method 1: Time Complexity：O(2^n). Space: O(n), stack space.

Method 2: Time Complexity: O(n). Space: O(n).

Method 3: Time Complexity: O(n). Space: O(1).

AC Java:

 1 public class Solution {
 2     public int climbStairs(int n) {
 3         /*
 4         //Method 1
 5         if(n == 0){
 6             return 1;
 7         }
 8         if(n == 1){
 9             return 1;
10         }
11         return climbStairs(n-1) + climbStairs(n-2);
12         */
13         
14         /*
15         //Method 2
16         int [] arr = new int[n+1];
17         arr[0] = 1;
18         arr[1] = 1;
19         for(int i = 2; i <= n; i++){
20             arr[i] = arr[i-1] + arr[i-2];
21         }
22         return arr[n];
23         */
24         //Method 3
25         if(n == 1){
26             return 1;
27         }
28         if(n == 2){
29             return 2;
30         }
31         int first = 1;
32         int second = 2;
33         int res = 0;
34         for(int i = 3; i<=n; i++){
35             res = first + second;
36             first = second;
37             second = res;
38         }
39         return res;
40     }
41 }

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