[LeetCode] 70. Climbing Stairs

You are climbing a staircase. 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

Constraints:

1 <= n <= 45

爬楼梯。

假设你正在爬楼梯。需要 n 阶你才能到达楼顶。

每次你可以爬 1 或 2 个台阶。你有多少种不同的方法可以爬到楼顶呢？

注意：给定 n 是一个正整数。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/climbing-stairs
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这道题有两种解法，一种是数学解法，一种是动态规划。

首先是数学解法，这个题本质上是斐波那契数列。当输入为1, 2, 3, 4, 5, 6, 7, 8, 9, 10时，观察输出为1, 2, 3, 5, 8, 13, 21, 34, 55, 89。所以做法就很直观了。

迭代

时间O(n)

空间O(n)

JavaScript实现

 1 /**
 2  * @param {number} n
 3  * @return {number}
 4  */
 5 var climbStairs = function (n) {
 6     const climing = [];
 7     // using variables because they allocate a memory only once
 8     let minusTwoSteps = 1;
 9     if (n === 1) {
10         return 1;
11     }
12     let minusOneStep = 2;
13     if (n === 2) {
14         return 2;
15     }
16     let current = 0;
17     for (let i = 3; i <= n; i++) {
18         current = minusTwoSteps + minusOneStep; // current step - is a sum of two previous ones
19         minusTwoSteps = minusOneStep; // -2 steps for next iteration would be current - first
20         minusOneStep = current; // -1 step for next iteration would be our current
21     }
22     return current;
23 };

Java实现

 1 class Solution {
 2     public int climbStairs(int n) {
 3         int first = 1;
 4         int second = 2;
 5         if (n <= 2) {
 6             return n;
 7         }
 8         int cur = 0;
 9         for (int i = 3; i <= n; i++) {
10             cur = first + second;
11             first = second;
12             second = cur;
13         }
14         return cur;
15     }
16 }

dp[i] 数组的定义是跑到第 i 层楼的时候，上楼梯的组合数是多少。几个初始值是 dp[0] = 0, dp[1] = 1, dp[2] = 2。因为每次既可以爬一层楼，也可以爬两层楼，所以当你需要知道第i层楼的爬法的时候，你需要看的是我爬到 i - 2 层楼有几种爬法和我爬到 i - 1 层楼有几种爬法。所以状态转移方程就是 dp[i] = dp[i - 1] + dp[i - 2]。

时间O(n)

空间O(n)

JavaScript实现

 1 /**
 2  * @param {number} n
 3  * @return {number}
 4  */
 5 var climbStairs = function (n) {
 6     if (n === 0) return 0;
 7     let dp = [n + 1];
 8     dp[0] = 1;
 9     dp[1] = 1;
10     for (let i = 2; i <= n; i++) {
11         dp[i] = dp[i - 1] + dp[i - 2];
12     }
13     return dp[n];
14 };

Java实现

 1 class Solution {
 2     public int climbStairs(int n) {
 3         if (n == 0) return 0;
 4         int[] dp = new int[n + 1];
 5         dp[0] = 1;
 6         dp[1] = 1;
 7         for (int i = 2; i <= n; i++) {
 8             dp[i] = dp[i - 1] + dp[i - 2];
 9         }
10         return dp[n];
11     }
12 }