LeetCode 70. Climbing Stairs

原题链接在这里：https://leetcode.com/problems/climbing-stairs/

题目：

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.

题解：

其实是Fibonacci Number. e.g. n = 100, 假设登到99台阶有m种方法，登到98台阶有n种方法，那么从99到100都是上一步，所以还是m种方法. 从98台阶登到100都是一次登两个台阶，还是n种方法，若果在98登一个台阶，就是到了99, 这种方法已经包含在最初登到99台阶的m种方法中了。

所以登到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:

public class Solution {
    public int climbStairs(int n) {
        /*
        //Method 1
        if(n == 0){
            return 1;
        }
        if(n == 1){
            return 1;
        }
        return climbStairs(n-1) + climbStairs(n-2);
        */
        
        /*
        //Method 2
        int [] arr = new int[n+1];
        arr[0] = 1;
        arr[1] = 1;
        for(int i = 2; i <= n; i++){
            arr[i] = arr[i-1] + arr[i-2];
        }
        return arr[n];
        */
        //Method 3
        if(n == 1){
            return 1;
        }
        if(n == 2){
            return 2;
        }
        int first = 1;
        int second = 2;
        int res = 0;
        for(int i = 3; i<=n; i++){
            res = first + second;
            first = second;
            second = res;
        }
        return res;
    }
}

类似Min Cost Climbing Stairs, New 21 Game, Domino and Tromino Tiling, Combination Sum IV.