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| // @ts-nocheck | ||
| /** | ||
| * 题目名称：爬楼梯（跳台阶） | ||
| * leetcode 题目: https://leetcode-cn.com/problems/climbing-stairs/ | ||
| * leetcode 题解: https://leetcode-cn.com/problems/climbing-stairs/solution/by-hovinghuang-bl5r/ | ||
| * 牛客网 题目: https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4?tpId=295&tqId=23261&ru=/exam/oj&qru=/ta/format-top101/question-ranking&sourceUrl=%2Fexam%2Foj%3Ftab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D295 | ||
| * 牛客网 题解: https://blog.nowcoder.net/n/7057ccebe8fd452580442b74ffbb8d90 | ||
| */ | ||
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| /** | ||
| * 解法一：递归（超时） | ||
| * 思路：满足斐波那契数列公式，最简单的肯定是递归 | ||
| * 时间复杂度：O(2^n) | ||
| * 空间复杂度： | ||
| */ | ||
| function climbStairs(n: number): number { | ||
| if (n <= 2) return n | ||
| return climbStairs(n - 1) + climbStairs(n - 2) | ||
| }; | ||
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| /** | ||
| * 解法二：循环（记忆化累加） | ||
| * 思路：n1、n2 记录前面两位的结果，一个循环搞定 | ||
| * 时间复杂度：O(n) | ||
| * 空间复杂度：O(1) | ||
| */ | ||
| function climbStairs(n: number): number { | ||
| if (n <= 2) return n | ||
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| let n1 = 1 // 记录 n - 1 的结果 | ||
| let n2 = 1 // 记录 n - 2 的结果 | ||
| let res = 0 | ||
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| for (let i = 2; i <= n; i++) { | ||
| res = n1 + n2 | ||
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| // 记录中间结果 | ||
| n2 = n1 | ||
| n1 = res | ||
| } | ||
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| return res | ||
| }; | ||
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| /** | ||
| * 解法三：动态规划 | ||
| * 思路： | ||
| * 时间复杂度：O(n) | ||
| * 空间复杂度： | ||
| */ | ||
| function climbStairs(n: number): number { | ||
| const dp = new Array(n + 1) | ||
| dp[1] = 1 | ||
| dp[2] = 2 | ||
| for (let i = 3; i <= n; i++){ | ||
| dp[i] = dp[i - 1] + dp[i - 2] | ||
| } | ||
| return dp[n] | ||
| }; |