计算覆盖距离的方法数| 2套

给定距离N 。的任务是计数的方法来覆盖1,2和3步的距离的总数。
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，讨论了基于递归和动态编程的方法。在这里，我们将降低空间复杂度。可以肯定的是，要计算覆盖距离i的步数，只需要最后三个状态( i – 1 ， i – 2 ， i – 3 )。因此，可以使用最后三个状态计算结果。
下面是上述方法的实现：

C++

// 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

Javascript

    // Javascript 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
    function countWays(n)
    {
        // Base conditions
        if (n == 0)
            return 1;
        if (n <= 2)
            return n;
 
        // To store the last three stages
        let f0 = 1, f1 = 1, f2 = 2;
        let ans = 0;
 
        // Find the numbers of steps required
        // to reach the distance i
        for (let i = 3; i <= n; i++)
        {
            ans = f0 + f1 + f2;
            f0 = f1;
            f1 = f2;
            f2 = ans;
        }
 
        // Return the required answer
        return ans;
    }
     
    let n = 4;
    document.write(countWays(n));
 
// This code is contributed by suresh07.

输出：

7

时间复杂度： O(N)
空间复杂度O(1)

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