打印所有到达第 N 级楼梯的方法，一次跳跃 1 或 2 个单位

给定一个代表N个楼梯的正整数N和一个人它在第一个楼梯，任务是打印所有到达第N^个楼梯的路径，一次跳跃1 或 2 个单位。

例子：

Input: N = 3
Output: 
11
2
Explanation:
Nth stairs can be reached in the following ways with the jumps of 1 or 2 units each as:

1. Perform the two jumps of 1 unit each as 1 -> 1.
2. Perform the two jumps of 1 unit each as 2.

Input: N = 5
Output:
1111
112
121
211
22

方法：给定的问题可以使用递归来解决。这个想法是在每个索引处同时涵盖一次或两次跳跃的情况，并打印到达第N^个楼梯的所有可能方式。请按照以下步骤解决给定的问题：

• 定义一个递归函数，比如totalPossibleJumps(int N) ，它返回所有可能的跳转方式以到达第N^个楼梯。
  • 在递归检查的起点，基本情况如下：
    • 如果N < 0 ，那么它不是一个有效的方法。所以返回一个空数组
    • 如果N = 0 ，那么它是一种有效的方式。所以返回一个包含一个空格的大小为1的数组。
  • 每次递归调用递归函数两次，一次为1 个单元跳转，另一次为2 个单元跳转，并分别存储结果。
  • 初始化一个字符串数组，比如totalJumps来存储到达第i^个索引的方式，并存储到达索引i的所有可能方式，结果存储在上述两个递归调用中。
• 完成上述步骤后，将上述递归调用返回的所有可能组合打印为totalPossibleJumps(N) 。

下面是上述方法的实现：  

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find all the ways to reach
// Nth stair using one or two jumps
vector TotalPossibleJumps(int N)
{
    // Base Cases
    if ((N - 1) == 0) {
        vector newvec;
        newvec.push_back("");
        return newvec;
    }
    else {
        if (N < 0) {
            vector newvec;
            return newvec;
        }
    }
 
    // Recur for jump1 and jump2
    vector jump1
        = TotalPossibleJumps(N - 1);
    vector jump2
        = TotalPossibleJumps(N - 2);
 
    // Stores the total possible jumps
    vector totaljumps;
 
    // Add "1" with every element
    // present in jump1
    for (string s : jump1) {
        totaljumps.push_back("1" + s);
    }
 
    // Add "2" with every element
    // present in jump2
    for (string s : jump2) {
        totaljumps.push_back("2" + s);
    }
 
    return totaljumps;
}
 
// Driver Code
int main()
{
    int N = 3;
    vector Ans = TotalPossibleJumps(N);
    for (auto& it : Ans)
        cout << it << '\n';
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find all the ways to reach
    // Nth stair using one or two jumps
    static ArrayList TotalPossibleJumps(int N)
    {
        // Base Cases
        if ((N - 1) == 0) {
            ArrayList newvec
                = new ArrayList();
            newvec.add("");
            return newvec;
        }
        else {
            if (N < 0) {
                ArrayList newvec
                    = new ArrayList();
                return newvec;
            }
        }
 
        // Recur for jump1 and jump2
        ArrayList jump1 = TotalPossibleJumps(N - 1);
        ArrayList jump2 = TotalPossibleJumps(N - 2);
 
        // Stores the total possible jumps
        ArrayList totaljumps
            = new ArrayList();
 
        // Add "1" with every element
        // present in jump1
        for (String s : jump1) {
            totaljumps.add("1" + s);
        }
 
        // Add "2" with every element
        // present in jump2
        for (String s : jump2) {
            totaljumps.add("2" + s);
        }
 
        return totaljumps;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 3;
        ArrayList Ans = TotalPossibleJumps(N);
        for (String it : Ans)
            System.out.println(it);
    }
}
 
// This code is contributed by ukasp.

Python3

# python program for the above approach
 
# Function to find all the ways to reach
# Nth stair using one or two jumps
 
 
def TotalPossibleJumps(N):
 
    # Base Cases
    if ((N - 1) == 0):
        newvec = []
        newvec.append("")
        return newvec
 
    else:
        if (N < 0):
            newvec = []
            return newvec
 
    # Recur for jump1 and jump2
    jump1 = TotalPossibleJumps(N - 1)
    jump2 = TotalPossibleJumps(N - 2)
 
    # Stores the total possible jumps
    totaljumps = []
 
    # Add "1" with every element
    # present in jump1
    for s in jump1:
        totaljumps.append("1" + s)
 
    # Add "2" with every element
    # present in jump2
    for s in jump2:
        totaljumps.append("2" + s)
 
    return totaljumps
 
 
# Driver Code
if __name__ == "__main__":
 
    N = 3
    Ans = TotalPossibleJumps(N)
    for it in Ans:
        print(it)
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find all the ways to reach
// Nth stair using one or two jumps
static List TotalPossibleJumps(int N)
{
    // Base Cases
    if ((N - 1) == 0) {
        List newvec = new List();
        newvec.Add("");
        return newvec;
    }
    else {
        if (N < 0) {
            List newvec = new List();
            return newvec;
        }
    }
 
    // Recur for jump1 and jump2
    List jump1
        = TotalPossibleJumps(N - 1);
    List jump2
        = TotalPossibleJumps(N - 2);
 
    // Stores the total possible jumps
    List totaljumps = new List();
 
    // Add "1" with every element
    // present in jump1
    foreach (string s in jump1) {
        totaljumps.Add("1" + s);
    }
 
    // Add "2" with every element
    // present in jump2
    foreach (string s in jump2) {
        totaljumps.Add("2" + s);
    }
 
    return totaljumps;
}
 
// Driver Code
public static void Main()
{
    int N = 3;
    List Ans = TotalPossibleJumps(N);
    foreach(string it in Ans)
        Console.WriteLine(it);
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

    // JavaScript program for the above approach
 
    // Function to find all the ways to reach
    // Nth stair using one or two jumps
    const TotalPossibleJumps = (N) => {
        // Base Cases
        if ((N - 1) == 0) {
            let newvec = [];
            newvec.push("");
            return newvec;
        }
        else {
            if (N < 0) {
                let newvec = [];
                return newvec;
            }
        }
 
        // Recur for jump1 and jump2
        let jump1 = TotalPossibleJumps(N - 1);
        let jump2 = TotalPossibleJumps(N - 2);
 
        // Stores the total possible jumps
        let totaljumps = [];
 
        // Add "1" with every element
        // present in jump1
        for (let s in jump1) {
            totaljumps.push("1" + jump1[s]);
        }
 
        // Add "2" with every element
        // present in jump2
        for (let s in jump2) {
            totaljumps.push("2" + jump2[s]);
        }
 
        return totaljumps;
    }
 
    // Driver Code
 
    let N = 3;
    let Ans = TotalPossibleJumps(N);
    for (let it in Ans)
        document.write(`${Ans[it]}<br/>`);
 
// This code is contributed by rakeshsahni

输出：

11
2

时间复杂度： O(2 ^N )
辅助空间： O(N)