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? 解决方案 方法一:暴力法 算法 在暴力法中,我们将会把所有可能爬的阶数进行组合,也就是 1 和 2 。而在每一步中我们都会继续调用 climbStairsclimbStairs 这个函数模拟爬 11 阶和22 阶的情...
Can you solve this real interview question? Climbing Stairs - 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
Climbing Stairs leetcode Climbing Stairs You are climbing a stair case. It takes n steps to reach to the top. 递归法 说明 state: f[i] 表示爬到第i个楼梯的所有方法的和 function: f[i] = f[i-1] + f[i-2] //因为每次走一步或者两步, 所以f[i]的方法就是它一步前和两步前方法加...
func climbStairs(n int) int { // dp[i] 表示上到第 i 级楼梯的方法数 dp := make([]int, n + 1) // 最开始在第 0 级和第 1 级各有一种方案 dp[0], dp[1] = 1, 1 for i := 2; i <= n; i++ { // 第 i 级可以从第 i - 1 级上一级到达, // 也可以从第 i - 2 ...
70.ClimbingStairsYou areclimbinga stair case. It takesnstepstoreachtothetop. Each time you can either climb1or2steps. In how many distinct ways can you climbtothetop? Note: Givenn LeetCode—Python—70. 爬楼梯 一、题目描述 斐波那契数列(70):https://leetcode.com/problems/climbing-stairs/des...
Leetcode 70. Climbing Stairs 爬楼梯 Leetcode 70. Climbing Stairs 爬楼梯 标签: Leetcode 题目地址: https://leetcode-cn.com/problems/climbing-stairs/ 题目描述 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 ...
LeetCode——70. Climbing Stairs 70. Climbing Stairs 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 ......
LeetCode官方题解方法一:动态规划 LeetCode精选题解第二种思路 复杂度分析 时间复杂度:O(N)O(N), 总计循环n次 空间复杂度:O(1)O(1), 采用滚动数组时只需要固定的额外空间 代码实现 Java版(滚动数组) classSolution{publicintclimbStairs(intn){int[] scrollArr = {0,0,1};// 滚动数组,节约空间for(...
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 positiv...leetcode -- 70. Climbing Stairs 题目描述 题目难度:Easy You are climbing a stair case. It takes n steps to reach to the top. Each time you can...
publicclassSolution {publicintclimbStairs(intn) {int[] memo =newint[n + 1];returnhelper(n, memo); }publicinthelper(intn,int[] memo) {if(n <= 1)return1;if(memo[n] > 0)returnmemo[n];returnmemo[n] = helper(n - 1, memo) + helper(n - 2, memo); ...