计算覆盖距离的方法数量|套装2

给定距离N。任务是计算以1步， 2步和3步覆盖距离的方法总数。

例子：

Input: N = 3
Output: 4
All the required ways are (1 + 1 + 1), (1 + 2), (2 + 1) and (3).

Input: N = 4
Output: 7

方法：在上一篇文章中，讨论了一种基于递归和动态编程的方法。在这里，我们将减少空间的复杂性。可以保留的是，要计算覆盖距离i的步数，仅需要最后三个状态( i – 1，i – 2，i – 3 )。因此，可以使用最后三个状态来计算结果。

下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count of the
// total number of ways to cover the
// distance with 1, 2 and 3 steps
int countWays(int n)
{
    // Base conditions
    if (n == 0)
        return 1;
    if (n <= 2)
        return n;
  
    // To store the last three stages
    int f0 = 1, f1 = 1, f2 = 2, ans;
  
    // Find the numbers of steps required
    // to reach the distance i
    for (int i = 3; i <= n; i++) {
        ans = f0 + f1 + f2;
        f0 = f1;
        f1 = f2;
        f2 = ans;
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
int main()
{
    int n = 4;
  
    cout << countWays(n);
  
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
// Function to return the count of the
// total number of ways to cover the
// distance with 1, 2 and 3 steps
static int countWays(int n)
{
    // Base conditions
    if (n == 0)
        return 1;
    if (n <= 2)
        return n;
  
    // To store the last three stages
    int f0 = 1, f1 = 1, f2 = 2;
    int ans=0;
  
    // Find the numbers of steps required
    // to reach the distance i
    for (int i = 3; i <= n; i++) 
    {
        ans = f0 + f1 + f2;
        f0 = f1;
        f1 = f2;
        f2 = ans;
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
public static void main (String[] args) 
{
      
    int n = 4;
    System.out.println (countWays(n));
}
}
  
// This code is contributed by jit_t

Python

# Python3 implementation of the approach
  
# Function to return the count of the
# total number of ways to cover the
# distance with 1, 2 and 3 steps
def countWays(n):
      
    # Base conditions
    if (n == 0):
        return 1
    if (n <= 2):
        return n
  
    # To store the last three stages
    f0 = 1
    f1 = 1
    f2 = 2
    ans = 0
  
    # Find the numbers of steps required
    # to reach the distance i
    for i in range(3, n + 1):
        ans = f0 + f1 + f2
        f0 = f1
        f1 = f2
        f2 = ans
  
    # Return the required answer
    return ans
  
# Driver code
n = 4
  
print(countWays(n))
  
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;
  
class GFG 
{
      
// Function to return the count of the
// total number of ways to cover the
// distance with 1, 2 and 3 steps
static int countWays(int n)
{
    // Base conditions
    if (n == 0)
        return 1;
    if (n <= 2)
        return n;
  
    // To store the last three stages
    int f0 = 1, f1 = 1, f2 = 2;
    int ans = 0;
  
    // Find the numbers of steps required
    // to reach the distance i
    for (int i = 3; i <= n; i++) 
    {
        ans = f0 + f1 + f2;
        f0 = f1;
        f1 = f2;
        f2 = ans;
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
public static void Main(String[] args) 
{
    int n = 4;
    Console.WriteLine (countWays(n));
}
}
  
// This code is contributed by PrinciRaj1992

输出：

7

时间复杂度： O(N)
空间复杂度O(1)