[LeetCode] 70. Climbing Stairs

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mb5fdb0f93c5ca2 2020-02-13 02:53:00

文章标签 array dynamic programming leetcode java javascript 文章分类 后端开发

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

爬楼梯。题意是给一个数字n代表楼梯的高度，你可以每次爬一步或者两步，求有多少种不同的爬法。

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)

 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 };

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 }