[LeetCode]70. Climbing Stairs

70. Climbing Stairs

一、题目

Problem Description:
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:
$1<=n<=45$

二、题解

• 爬楼梯问题是一道经典的DP问题
• 第n个楼梯可以由第n-1楼梯或第n-2个楼梯爬上来
• 其实这道爬楼问题就是经典的斐波那契数列问题：F(0)=0，F(1)=1, F(n)=F(n - 1)+F(n - 2)
• 由于题目中n>=0，则初始条件给定F(1)=1，F(2)=2
  

2.1 Approach #1 : Recursion / Brute Force

注：Wrong:Time Limit Exceeded
以下代码由于时间复杂度过高，无法通过

Time complexity : $O(2^n)$. Size of recursion tree will be $2^n$.
Recursion tree for n=5 would be like this:

Space complexity : $O(n)$. The depth of the recursion tree can go upto $n$.

优点：代码简洁明了，可读性高
缺点：如上图二叉树所示，存在大量重复计算

//Recursion / Brute Force
//Time complexity : O(2^n); Space complexity : O(n)
class Solution {
    public int climbStairs(int n) {
		return F(n);
    }
	
	public int F(int n){
		if(n == 1) return 1;
		if(n == 2) return 2;
		return F(n-1) + F(n-2);
	}
}

2.2 Approach #2 : Dynamic Programming

空间换时间：dp[]数组用来存储各个F(n)值

Time complexity : $O(n)$
Space complexity : $O(n)$

//Dynamic Programming
//Time complexity : O(n); Space complexity : O(n)
class Solution {
    public int climbStairs(int n) {
		if(n == 1) return 1;
		int[] dp = new int[n+1];
		dp[1] = 1;
		dp[2] = 2;
		for(int i = 3; i <= n; i++){
			dp[i] = dp[i-1] + dp[i-2];
		}
		return dp[n]; 
    }
}

2.3 Approach #3 : Optimization of Dynamic Programming

无需dp[]数组，只需两个变量保存F(n-1)和F(n-2)值即可

Time complexity : $O(n)$
Space complexity : $O(1)$

//Dynamic Programming
//Time complexity : O(n); Space complexity : O(1)
class Solution {
    public int climbStairs(int n) {
		if(n == 1) return 1;
		int num2 = 1;//num2表示F(n-2);num2初始化为F(1)=1
		int num1 = 2;//num1表示F(n-1);num1初始化为F(2)=2
		for(int i = 3; i <= n; i++){
			int num = num1 + num2;//num表示F(n);F(n)=F(n-2)+F(n-1)
			num2 = num1;
			num1 = num;
		}
		return num1;
    }
}