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;...
【动态规划】Leetcode编程题解:70. Climbing Stairs 题目: 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?提示:Given n will be a positive integer....
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 either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note: Given n will ...[LeetCode]70. Climbing Stairs 题目Solution......
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(...
LeetCode:Climbing Stairs 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? 1classSolution {2public:3intclimbStairs(intn) {4if(n==0)5return0;6elseif(n==1)7...
【leetcode】Climbing Stairs 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? 题解,变相的斐波那契数列。因为最后一步到达n的时候有两种选择,走1步走到台阶n和走2步走到...
Solution:一个台阶的方法次数为1次,两个台阶的方法次数为2个。n个台阶的方法可以理解成上n-2个台阶,然后2步直接上最后一步;或者上n-1个台阶,再单独上一步 1intclimbStairs(intn) {//Runtime:0ms2inta=1,b=2;3if(n==1)returna;4if(n==2)returnb;5intc=0;6for(inti=3;i<=n;i++){7c=a+...
代码: classSolution {public:intclimbStairs(intn) {if(n <=2)returnn;intstep =3, s1 =1, s2 =2, tmp;while(step <=n) { tmp= s1 +s2; s1=s2; s2=tmp;++step; }returns2; } }; 递归法(在LeetCode上会超时,应该是堆栈大小太大了): ...
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? 其实就是斐波那契数列,递推求解便知。 1classSolution {2public:3intclimbStairs(intn) {4if(n==1)return1;5if(n==2...