爬楼梯(Climbing Stairs) 70. 爬楼梯 70. 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 是一个正整数。 LeetCode70. Climbing Stairs 示例 1: 输入: 2 输出: 2 解释: 有两种方法可以爬到...
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...
public int climbStairs(int n) { if(n==1)return 1; else if(n==2)return 2; else{ return climbStairs(n-1)+climbStairs(n-2); } } } 解法二(非递归): class Solution { public int climbStairs(int n) { if(n==1)return 1; else if(n==2)return 2; int x=1,y=2; int res=0;...
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
代码参考:https://leetcode.com/problems/climbing-stairs/discuss/25315/My-DP-solution-in-C%2B%2B-with-explanation 2.3 自下向上的动态规划算法 具体代码如下: classSolution{public:intclimbStairs(intn){if(n ==1|| n ==2) {returnn; }intone_step_before =2;inttwo_steps_before =1;intnow_steps...
class Solution { public: int climbStairs(int n) { if(n == 1) return 1; if(n == 2) return 2; int f = 1, s = 2, res = 0; for(int i = 3; i <= n; ++i){ res = f + s; f = s; s = res; } return res; } }; // class Solution { // public: // int climb...
Leetcode:Climbing Stairs 开始考虑排列组合方法,发现自己实现的求组合函数C(m,n)过大数据时有溢出,换了思路。 1intclimbStairs(intn) {2//Start typing your C/C++ solution below3//DO NOT write int main() function4int* res=newint[n];5res[0]=1;6res[1]=2;7for(inti =2; i<n;i++){8...
【Leetcode】Climbing Stairs 题目: You are climbing a stair case. It takesnsteps 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? 解题思路:这是一道斐波拉契数列题。看到斐波那契数列就自然想到利用递归,然而在这里递归的...
LeetCode-Climbing Stairs 斐波那契数列; 1classSolution {2public:3intclimbStairs(intn) {4//Start typing your C/C++ solution below5//DO NOT write int main() function6if(n <=0) {7return0;8}9vector<int> fb(n +1,1);10fb[2] =2;11for(inti =3; i <= n; ++i) {12fb[i] = fb...
leetcode-climbing stairs 第一种是迭代,第二种是DP,第三种是斐波那契数列的通项公式。 此外,soulmachine的书上还指出,递归也可以,只是速度太慢。 1#include <iostream>2#include <cmath>3usingnamespacestd;45classSolution {6public:7intclimbStairs(intn) {8intprev =0;9intcur =1;10for(inti =1; i ...