计算覆盖距离的方法数

给定一个距离 ‘dist，计算用 1、2 和 3 步覆盖该距离的方法总数。

例子：

Input: n = 3
Output: 4
Explanation:
Below are the four ways
 1 step + 1 step + 1 step
 1 step + 2 step
 2 step + 1 step
 3 step

Input: n = 4
Output: 7
Explanation:
Below are the four ways
 1 step + 1 step + 1 step + 1 step
 1 step + 2 step + 1 step
 2 step + 1 step + 1 step 
 1 step + 1 step + 2 step
 2 step + 2 step
 3 step + 1 step
 1 step + 3 step

递归解

• 方法：有n个楼梯，一个人可以走下一步，跳过一个位置或跳过两个位置。所以有n个位置。这个想法是站在人可以移动 i+1、i+2、i+3 位置的第 i 个位置。因此可以形成一个递归函数，其中在当前索引 i 处，该函数被递归调用到 i+1、i+2 和 i+3 位置。
  还有另一种形成递归函数。要到达位置 i，一个人必须从 i-1、i-2 或 i-3 位置跳跃，其中 i 是起始位置。
• 算法：
  1. 创建一个递归函数( count(int n) )，它只接受一个参数。
  2. 检查基本情况。如果 n 的值小于 0，则返回 0，如果 n 的值等于 0，则返回 1，因为它是起始位置。
  3. 使用值 n-1、n-2 和 n-3 递归调用该函数并对返回的值求和，即sum = count(n-1) + count(n-2) + count(n-3) 。
  4. 返回sum的值。
• 执行：

C++

// A naive recursive C++ program to count number of ways to cover
// a distance with 1, 2 and 3 steps
#include
using namespace std;
 
// Returns count of ways to cover 'dist'
int printCountRec(int dist)
{
    // Base cases
    if (dist<0)      return 0;
    if (dist==0)  return 1;
 
    // Recur for all previous 3 and add the results
    return printCountRec(dist-1) +
           printCountRec(dist-2) +
           printCountRec(dist-3);
}
 
// driver program
int main()
{
    int dist = 4;
    cout << printCountRec(dist);
    return 0;
}

Java

// A naive recursive Java program to count number
// of ways to cover a distance with 1, 2 and 3 steps
import java.io.*;
 
class GFG
{
    // Function returns count of ways to cover 'dist'
    static int printCountRec(int dist)
    {
        // Base cases
        if (dist<0)   
            return 0;
        if (dist==0)   
            return 1;
  
        // Recur for all previous 3 and add the results
        return printCountRec(dist-1) +
               printCountRec(dist-2) +
               printCountRec(dist-3);
    }
     
    // driver program
    public static void main (String[] args)
    {
        int dist = 4;
        System.out.println(printCountRec(dist));
    }
}
 
// This code is contributed by Pramod Kumar

Python3

# A naive recursive Python3 program
# to count number of ways to cover
# a distance with 1, 2 and 3 steps
 
# Returns count of ways to
# cover 'dist'
def printCountRec(dist):
     
    # Base cases
    if dist < 0:
        return 0
         
    if dist == 0:
        return 1
 
    # Recur for all previous 3 and      
   # add the results
    return (printCountRec(dist-1) +
            printCountRec(dist-2) +
            printCountRec(dist-3))
 
# Driver code
dist = 4
print(printCountRec(dist))
# This code is contributed by Anant Agarwal.

C#

// A naive recursive C# program to
// count number of ways to cover a
// distance with 1, 2 and 3 steps
using System;
 
class GFG {
     
    // Function returns count of
    // ways to cover 'dist'
    static int printCountRec(int dist)
    {
        // Base cases
        if (dist < 0)
            return 0;
        if (dist == 0)
            return 1;
 
        // Recur for all previous 3
        // and add the results
        return printCountRec(dist - 1) +
               printCountRec(dist - 2) +
               printCountRec(dist - 3);
    }
     
    // Driver Code
    public static void Main ()
    {
        int dist = 4;
        Console.WriteLine(printCountRec(dist));
    }
}
 
// This code is contributed by Sam007.

Javascript

 
// A naive recursive javascript program to count number of ways to cover
// a distance with 1, 2 and 3 steps
 
// Returns count of ways to cover 'dist'
function printCountRec(dist)
{
    // Base cases
    if (dist<0)     return 0;
    if (dist==0) return 1;
 
    // Recur for all previous 3 and add the results
    return printCountRec(dist-1) +
        printCountRec(dist-2) +
        printCountRec(dist-3);
}
 
// driver program
var dist = 4;
document.write(printCountRec(dist));
 

输出：

7

• 复杂度分析：
  • 时间复杂度： O(3 ⁿ )。
    上述解决方案的时间复杂度是指数级的，接近的上限是 O(3 ⁿ )。从每个状态 3 调用一个递归函数。所以 n 个状态的上限是 O(3 ⁿ )。
  • 空间复杂度： O(1)。
    不需要额外的空间。

高效的解决方案

• 做法：思路类似，但是可以观察到有n个状态但是递归函数被调用了3^n次。这意味着某些状态会被重复调用。所以这个想法是存储状态的值。这可以通过两种方式完成。
  • 第一种方法是保持递归结构完整，只将值存储在 HashMap 中，每当调用函数，不计算就返回值存储(自上而下方法)。
  • 第二种方式是取一个额外的n大小的空间，开始计算从1, 2 ..到n的状态值，即计算i, i+1, i+2的值，然后用它们来计算i的值+3(自下而上的方法)。
  • 动态规划中的重叠子问题。
  • 动态规划中的最优子结构属性。
  • 动态规划(DP)问题
• 算法：
  1. 创建一个大小为 n + 1 的数组，并用 1、1、2 初始化前 3 个变量。基本情况。
  2. 运行从 3 到 n 的循环。
  3. 对于每个索引 i，将第 i 个位置的值计算为dp[i] = dp[i-1] + dp[i-2] + dp[i-3] 。
  4. 打印 dp[n] 的值，作为覆盖一段距离的方法数。
• 执行：

C++

// A Dynamic Programming based C++ program to count number of ways
// to cover a distance with 1, 2 and 3 steps
#include
using namespace std;
 
int printCountDP(int dist)
{
    int count[dist+1];
 
    // Initialize base values. There is one way to cover 0 and 1
    // distances and two ways to cover 2 distance
     count[0] = 1;
     if(dist >= 1)
            count[1] = 1;
     if(dist >= 2)
              count[2] = 2;
 
    // Fill the count array in bottom up manner
    for (int i=3; i<=dist; i++)
       count[i] = count[i-1] + count[i-2] + count[i-3];
 
    return count[dist];
}
 
// driver program
int main()
{
    int dist = 4;
    cout << printCountDP(dist);
    return 0;
}

Java

// A Dynamic Programming based Java program
// to count number of ways to cover a distance
// with 1, 2 and 3 steps
import java.io.*;
 
class GFG
{
    // Function returns count of ways to cover 'dist'
    static int printCountDP(int dist)
    {
        int[] count = new int[dist+1];
  
        // Initialize base values. There is one way to
        // cover 0 and 1 distances and two ways to
        // cover 2 distance
        count[0] = 1;
          if(dist >= 1)
            count[1] = 1;
        if(dist >= 2)
              count[2] = 2;
  
        // Fill the count array in bottom up manner
        for (int i=3; i<=dist; i++)
            count[i] = count[i-1] + count[i-2] + count[i-3];
  
        return count[dist];
    }
     
    // driver program
    public static void main (String[] args)
    {
        int dist = 4;
        System.out.println(printCountDP(dist));
    }
}
 
// This code is contributed by Pramod Kumar

蟒蛇3

# A Dynamic Programming based on Python3
# program to count number of ways to
# cover a distance with 1, 2 and 3 steps
 
def printCountDP(dist):
    count = [0] * (dist + 1)
     
    # Initialize base values. There is
    # one way to cover 0 and 1 distances
    # and two ways to cover 2 distance
    count[0] = 1
    if dist >= 1 :
        count[1] = 1
    if dist >= 2 :
        count[2] = 2
     
    # Fill the count array in bottom
    # up manner
    for i in range(3, dist + 1):
        count[i] = (count[i-1] +
                   count[i-2] + count[i-3])
         
    return count[dist];
 
# driver program
dist = 4;
print( printCountDP(dist))
 
# This code is contributed by Sam007.

C＃

// A Dynamic Programming based C# program
// to count number of ways to cover a distance
// with 1, 2 and 3 steps
using System;
 
class GFG {
     
    // Function returns count of ways
    // to cover 'dist'
    static int printCountDP(int dist)
    {
        int[] count = new int[dist + 1];
 
        // Initialize base values. There is one
        // way to cover 0 and 1 distances
        // and two ways to cover 2 distance
        count[0] = 1;
        count[1] = 1;
        count[2] = 2;
 
        // Fill the count array
        // in bottom up manner
        for (int i = 3; i <= dist; i++)
            count[i] = count[i - 1] +
                       count[i - 2] +
                       count[i - 3];
 
        return count[dist];
    }
     
    // Driver Code
    public static void Main ()
    {
        int dist = 4;
        Console.WriteLine(printCountDP(dist));
    }
}
 
// This code is contributed by Sam007.

Javascript

 
// A Dynamic Programming based Javascript program
// to count number of ways to cover a distance
// with 1, 2 and 3 steps
   
// Function returns count of ways
// to cover 'dist'
function printCountDP(dist)
{
    let count = new Array(dist + 1);
     
    // Initialize base values. There is one
    // way to cover 0 and 1 distances
    // and two ways to cover 2 distance
    count[0] = 1;
    if (dist >= 1)
        count[1] = 1;
    if (dist >= 2)
        count[2] = 2;
 
    // Fill the count array
    // in bottom up manner
    for(let i = 3; i <= dist; i++)
        count[i] = count[i - 1] +
                   count[i - 2] +
                   count[i - 3];
 
    return count[dist];
}
 
// Driver code
let dist = 4;
document.write(printCountDP(dist));
   
// This code is contributed by divyeshrabadiya07
 

输出 ：

7

• 复杂度分析：
  • 时间复杂度： O(n)。
    只需要遍历一次数组。所以时间复杂度是 O(n)
  • 空间复杂度： O(n)。
    将值存储在 DP O(n) 中需要额外的空间。

更优的解决方案

方法：代替使用大小为 n+1 的数组，我们可以使用大小为 3 的数组，因为为了计算特定步骤的方法数，我们只需要最后 3 步方法数。

算法：

1. 创建一个大小为 3 的数组并将步骤 0、1、2 的值初始化为 1、1、2(基本情况)。
2. 运行从 3 到 n(dist) 的循环。
3. 对于每个索引，将值计算为 way[i%3] = way[(i-1)%3] + way[(i-2)%3] + way[(i-3)%3] 并存储其值在数组方式的 i%3 索引处。如果我们计算索引 3 的值，则计算值将位于索引 0，因为对于较大的索引(4 ,5,6…..)，我们不需要索引 0 的值。
4. 返回 way[n%3] 的值。

C++

// A Dynamic Programming based C++ program to count number of ways
#include
using namespace std;
  
int printCountDP(int dist)
{
        //Create the array of size 3.
        int  ways[3] , n = dist;
         
        //Initialize the bases cases
        ways[0] = 1;
        ways[1] = 1;
        ways[2] = 2;
         
        //Run a loop from 3 to n
        //Bottom up approach to fill the array
        for(int i=3 ;i<=n ;i++)
            ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3];
         
        return ways[n%3];
}
  
// driver program
int main()
{
    int dist = 4;
    cout << printCountDP(dist);
    return 0;
}

Java

// A Dynamic Programming based Java program to count number of ways
import java.util.*;
  
class GFG {
  
static int printCountDP(int dist)
{
        // Create the array of size 3.
        int []ways = new int[3];
        int n = dist;
         
         
        // Initialize the bases cases
        ways[0] = 1;
        ways[1] = 1;
        ways[2] = 2;
         
        // Run a loop from 3 to n
        // Bottom up approach to fill the array
        for(int i=3 ;i<=n ;i++)
            ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3];
         
        return ways[n%3];
}
  
// driver program
public static void main(String arg[])
    {
    int dist = 4;
    System.out.print(printCountDP(dist));
    }
}
 
// this code is contributed by shivanisinghss2110

蟒蛇3

# A Dynamic Programming based C++ program to count number of ways
def prCountDP( dist):
 
        # Create the array of size 3.
        ways = [0]*3
        n = dist
         
        # Initialize the bases cases
        ways[0] = 1
        ways[1] = 1
        ways[2] = 2
         
        # Run a loop from 3 to n
        # Bottom up approach to fill the array
        for i in range(3, n + 1):
            ways[i % 3] = ways[(i - 1) % 3] + ways[(i - 2) % 3] + ways[(i - 3) % 3]
         
        return ways[n % 3]
  
# driver program
dist = 4
print(prCountDP(dist))
 
# This code is contributed by shivanisinghss2110

C＃

// A Dynamic Programming based C#
// program to count number of ways
using System;
  
class GFG{
  
static int printCountDP(int dist)
{
    // Create the array of size 3.
    int []ways = new int[3];
    int n = dist;
     
    // Initialize the bases cases
    ways[0] = 1;
    ways[1] = 1;
    ways[2] = 2;
     
    // Run a loop from 3 to n
    // Bottom up approach to fill the array
    for(int i = 3; i <= n; i++)
        ways[i % 3] = ways[(i - 1) % 3] +
                      ways[(i - 2) % 3] +
                      ways[(i - 3) % 3];
     
    return ways[n % 3];
}
  
// Driver code
public static void Main(String []arg)
{
    int dist = 4;
     
    Console.Write(printCountDP(dist));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

// A Dynamic Programming based javascript program to count number of ways
   
function printCountDP( dist)
{
        //Create the array of size 3.
        var  ways= [] , n = dist;
        ways.length = 3 ;
          
        //Initialize the bases cases
        ways[0] = 1;
        ways[1] = 1;
        ways[2] = 2;
          
        //Run a loop from 3 to n
        //Bottom up approach to fill the array
        for(var i=3 ;i<=n ;i++)
            ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3];
          
        return ways[n%3];
}
   
// driver code
 
    var dist = 4;
    document.write(printCountDP(dist));

输出 ：

7

时间复杂度：O(n)

空间复杂度：O(1)

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