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Java实战中的青蛙跳台阶问题
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Share interests, spread happiness, increase knowledge, and leave a good future!
Dear you,
this is LearningYard Academy
Today I bring you this
The frog on the stairs problem in Java practice
Welcome to visit!
青蛙跳台阶问题是经典的动态规划问题,通常被用来练习递推关系和动态规划的思维方式。
The frog on the stairs problem is a classic dynamic programming problem that is often used to practice recursive relationships and dynamic programming thinking.
1.问题描述
1. Problem description
一只青蛙要跳上一个有n阶台阶的楼梯,每次它可以跳1阶或2阶。请问,青蛙跳上n阶台阶的跳法有多少种?
A frog wants to jump up a staircase with n steps, each time it can jump 1 step or 2 steps. How many ways can the frog jump up n steps?
2.思路
2. Ideas
这个问题可以通过动态规划的思想来解决。假设dp[i]表示跳上第i阶台阶的跳法数。那么,青蛙跳上第i阶台阶的跳法可以由以下两种方式得到:
This problem can be solved by the idea of dynamic programming. Suppose dp[i] represents the number of ways to jump on the i-th step. Then, the frog can jump on the i-th step in one of two ways:
青蛙从第i-1阶跳到第i阶,即它已经跳到了i-1阶,然后再跳1阶。
The frog jumps from the i-1 step to the i step, that is, it has already jumped to the i-1 step and then one more step.
青蛙从第i-2阶跳到第i阶,即它已经跳到了i-2阶,然后再跳2阶。
The frog jumps from the i-2 step to the i step, that is, it has already jumped to the i-2 step and then jumps two more steps.
因此,状态转移方程为:dp[i] = dp[i-1] + dp[i-2]
Therefore, the state transition equation is: dp[i] = dp[i-1] + dp[i-2]
3.代码实现
3. Code implementation
以下是Java语言的实现:
The following is the implementation of Java language:
public class FrogJump {
public static int numWays(int n) {
if (n == 0) {
return 1; // 没有台阶,只有一种方式,即不跳
}
if (n == 1) {
return 1; // 只有一阶台阶,只有一种跳法
}
int dp = new int[n + 1];
dp[0] = 1; // 地面
dp[1] = 1; // 第1阶台阶
for (int i = 2; i
dp[i] = dp[i - 1] + dp[i - 2]; // 从i-1阶和i-2阶跳上来
}
return dp[n]; // 返回跳上n阶台阶的跳法数
}
public static void main(String args) {
int n = 5; // 举例
System.out.println("青蛙跳上" + n + "阶台阶的跳法数为: " + numWays(n));
}
}
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翻译来源于百度翻译
部分来源于百度文库
来源:牛人科技说