通过跳跃 1 到 N 来计算到达第 N 级楼梯的方式数

给定一个描述楼梯数量的整数N ，任务是计算通过 1 到 N 的跳跃到达第 N^个楼梯的方式的数量。

例子：

Input: N = 2
Output: 2
Explanation: Two ways to reach are: (1, 1) & (2)

Input: N = 3
Output: 4

Input: N = 4
Output: 8

方法：在这个问题中，到达第i^个楼梯的路径数是：

Ways to reach i^th stair = (Sum of ways to reach stairs 1 to i-1)+1
As for any stair before i, i^th stair can be reached in a single jump. And +1 for jumping directly to i.

现在解决这个问题：

1. 创建一个变量sum以存储到达特定楼梯的方式数。用0初始化它。
2. 运行从i=1到i=N-1的循环，每次迭代：
   1. 创建一个变量，比如cur来存储当前楼梯的路径数。所以， cur = sum + 1 。
   2. 将 sum 更改为sum = sum + cur 。
3. 循环结束后返回sum + 1作为此问题的答案。

下面是上述方法的实现：

C++

// C++ code for the above approach
 
#include 
using namespace std;
 
// Function to count the number of ways
// to reach Nth stair
int findWays(int N)
{
 
    int sum = 0;
    for (int i = 1; i < N; i++) {
        int cur = sum + 1;
        sum += cur;
    }
 
    return sum + 1;
}
 
// Driver Code
int main()
{
    int N = 10;
    cout << findWays(N);
}

Java

// Java code for the above approach
import java.util.*;
public class GFG
{
 
  // Function to count the number of ways
  // to reach Nth stair
  static int findWays(int N)
  {
 
    int sum = 0;
    for (int i = 1; i < N; i++) {
      int cur = sum + 1;
      sum += cur;
    }
 
    return sum + 1;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int N = 10;
    System.out.print(findWays(N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# python3 code for the above approach
 
# Function to count the number of ways
# to reach Nth stair
def findWays(N):
 
    sum = 0
    for i in range(1, N):
        cur = sum + 1
        sum += cur
 
    return sum + 1
 
# Driver Code
if __name__ == "__main__":
 
    N = 10
    print(findWays(N))
 
# This code is contributed by rakeshsahni

C#

// C# code to implement above approach
using System;
class GFG
{
 
  // Function to count the number of ways
  // to reach Nth stair
  static int findWays(int N)
  {
 
    int sum = 0;
    for (int i = 1; i < N; i++) {
      int cur = sum + 1;
      sum += cur;
    }
 
    return sum + 1;
  }
 
  // Driver code
  public static void Main()
  {
    int N = 10;
    Console.Write(findWays(N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

      // JavaScript code for the above approach
 
      // Function to count the number of ways
      // to reach Nth stair
      function findWays(N) {
 
          let sum = 0;
          for (let i = 1; i < N; i++) {
              let cur = sum + 1;
              sum += cur;
          }
 
          return sum + 1;
      }
 
      // Driver Code
 
      let N = 10;
      document.write(findWays(N));
 
// This code is contributed by Potta Lokesh
  

输出

512

时间复杂度： O(N)
辅助空间： O(1)