通过向右、向下或对角线移动从 M x N 矩阵的左上角到右下角的可能路径计数

给定 2 个整数M和N，任务是找到M x N矩阵的从左上角到右下角的所有可能路径的计数，约束条件是每个单元格只能向右或向下或对角线移动

例子：

Input:  M = 3, N = 3
Output: 13
Explanation: There are 13 paths as follows: VVHH, VHVH, HVVH, DVH, VDH, VHHV, HVHV, DHV, HHVV, HDV, VHD, HVD, and DD where V represents vertical, H represents horizontal, and D represents diagonal paths.

Input:  M = 2, N = 2
Output: 3

方法：这个想法是使用递归来找到路径的总数。这种方法与本文中讨论的方法非常相似。请按照以下步骤解决问题：

• 如果M或N等于1，则返回1。
• 否则，创建一个递归函数numberOfPaths() 为分别表示垂直、水平和对角线移动的值{M-1, N}、{M, N-1}和{M-1, N-1}调用相同的函数。

下面是上述方法的实现：

C++

// C++  program for the above approach
#include 
using namespace std;
 
// Returns count of possible paths to
// reach cell at row number M and column
// number N from the topmost leftmost
// cell (cell at 1, 1)
int numberOfPaths(int M, int N)
{
 
    // If either given row number or
    // given column number is first
    if (M == 1 || N == 1)
        return 1;
 
    // Horizontal Paths +
    // Vertical Paths +
    // Diagonal Paths
    return numberOfPaths(M - 1, N)
          + numberOfPaths(M, N - 1)
           + numberOfPaths(M - 1, N - 1);
}
 
// Driver Code
int main()
{
    cout << numberOfPaths(3, 3);
    return 0;
}

Java

// Java  program for the above approach
import java.util.*;
public class GFG
{
   
// Returns count of possible paths to
// reach cell at row number M and column
// number N from the topmost leftmost
// cell (cell at 1, 1)
static int numberOfPaths(int M, int N)
{
 
    // If either given row number or
    // given column number is first
    if (M == 1 || N == 1)
        return 1;
 
    // Horizontal Paths +
    // Vertical Paths +
    // Diagonal Paths
    return numberOfPaths(M - 1, N)
          + numberOfPaths(M, N - 1)
           + numberOfPaths(M - 1, N - 1);
}
 
// Driver Code
public static void main(String args[])
{
    System.out.print(numberOfPaths(3, 3));
 
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python

# Python  program for the above approach
 
# Returns count of possible paths to
# reach cell at row number M and column
# number N from the topmost leftmost
# cell (cell at 1, 1)
def numberOfPaths(M, N):
 
    # If either given row number or
    # given column number is first
    if (M == 1 or N == 1):
        return 1
 
    # Horizontal Paths +
    # Vertical Paths +
    # Diagonal Paths
    return numberOfPaths(M - 1, N) \
        + numberOfPaths(M, N - 1) \
        + numberOfPaths(M - 1, N - 1)
 
# Driver Code
print(numberOfPaths(3, 3))
 
# This code is contributed by Samim Hossain Mondal.

C#

// C#  program for the above approach
using System;
 
public class GFG
{
   
// Returns count of possible paths to
// reach cell at row number M and column
// number N from the topmost leftmost
// cell (cell at 1, 1)
static int numberOfPaths(int M, int N)
{
 
    // If either given row number or
    // given column number is first
    if (M == 1 || N == 1)
        return 1;
 
    // Horizontal Paths +
    // Vertical Paths +
    // Diagonal Paths
    return numberOfPaths(M - 1, N)
          + numberOfPaths(M, N - 1)
           + numberOfPaths(M - 1, N - 1);
}
 
// Driver Code
public static void Main(String []args)
{
    Console.Write(numberOfPaths(3, 3));
 
}
}
 
// This code is contributed by 29AjayKumar

Javascript

       // JavaScript code for the above approach
 
       // Returns count of possible paths to
       // reach cell at row number M and column
       // number N from the topmost leftmost
       // cell (cell at 1, 1)
       function numberOfPaths(M, N) {
 
           // If either given row number or
           // given column number is first
           if (M == 1 || N == 1)
               return 1;
 
           // Horizontal Paths +
           // Vertical Paths +
           // Diagonal Paths
           return numberOfPaths(M - 1, N)
               + numberOfPaths(M, N - 1)
               + numberOfPaths(M - 1, N - 1);
       }
 
       // Driver Code
       document.write(numberOfPaths(3, 3));
 
 // This code is contributed by Potta Lokesh
   

输出

13

时间复杂度： O(3 ^M*N )
辅助空间： O(1)

有效方法： Dp（使用记忆）

C++

// C++ program for the above approach
#include 
using namespace std;
 
static int dp[1001][1001];
 
// Returns count of possible paths to
// reach cell at row number M and column
// number N from the topmost leftmost
// cell (cell at 1, 1)
int numberOfPaths(int M, int N)
{
 
    // If either given row number or
    // given column number is first
    if (M == 1 || N == 1)
        return 1;
     
    // If a value already present
    // in t[][], return it
    if(dp[M][N] != -1) {
        return dp[M][N];
    }
     
    // Horizontal Paths +
    // Vertical Paths +
    // Diagonal Paths
    return dp[M][N] = numberOfPaths(M - 1, N)
                    + numberOfPaths(M, N - 1)
                    + numberOfPaths(M - 1, N - 1);
}
 
// Driver Code
int main()
{
    memset(dp,-1,sizeof(dp));
    cout << numberOfPaths(3, 3);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG {
  static int[][] dp = new int[1001][1001];
 
  // Returns count of possible paths to
  // reach cell at row number M and column
  // number N from the topmost leftmost
  // cell (cell at 1, 1)
  static int numberOfPaths(int M, int N)
  {
 
    // If either given row number or
    // given column number is first
    if (M == 1 || N == 1)
      return 1;
 
    // If a value already present
    // in t[][], return it
    if (dp[M][N] != -1) {
      return dp[M][N];
    }
 
    // Horizontal Paths +
    // Vertical Paths +
    // Diagonal Paths
    dp[M][N] = numberOfPaths(M - 1, N)
      + numberOfPaths(M, N - 1)
      + numberOfPaths(M - 1, N - 1);
    return dp[M][N];
  }
 
  // Driver Code
  public static void main(String args[])
  {
    for (int i = 0; i < 1001; i++)
      for (int j = 0; j < 1001; j++)
        dp[i][j] = -1;
    System.out.println(numberOfPaths(3, 3));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python

# Python program for the above approach
 
# Taking the matrix as globally
dp = [[-1 for i in range(1001)] for j in range(1001)]
 
# Returns count of possible paths to
# reach cell at row number M and column
# number N from the topmost leftmost
# cell (cell at 1, 1)
def numberOfPaths(M, N):
 
    # If either given row number or
    # given column number is first
    if (M == 1 or N == 1):
        return 1
 
    # If a value already present
    # in t[][], return it
    if(dp[M][N] != -1):
        return dp[M][N]
 
    # Horizontal Paths +
    # Vertical Paths +
    # Diagonal Paths
    dp[M][N] = numberOfPaths(M - 1, N) + numberOfPaths(M,
                     N - 1) + numberOfPaths(M - 1, N - 1)
    return dp[M][N]
 
# Driver Code
print(numberOfPaths(3, 3))
 
# This code is contributed by Samim Hossain Mondal.

C＃

// C# program for the above approach
using System;
class GFG {
    static int[, ] dp = new int[1001, 1001];
 
    // Returns count of possible paths to
    // reach cell at row number M and column
    // number N from the topmost leftmost
    // cell (cell at 1, 1)
    static int numberOfPaths(int M, int N)
    {
 
        // If either given row number or
        // given column number is first
        if (M == 1 || N == 1)
            return 1;
 
        // If a value already present
        // in t[][], return it
        if (dp[M, N] != -1) {
            return dp[M, N];
        }
 
        // Horizontal Paths +
        // Vertical Paths +
        // Diagonal Paths
        dp[M, N] = numberOfPaths(M - 1, N)
                   + numberOfPaths(M, N - 1)
                   + numberOfPaths(M - 1, N - 1);
        return dp[M, N];
    }
 
    // Driver Code
    public static void Main()
    {
        for (int i = 0; i < 1001; i++)
            for (int j = 0; j < 1001; j++)
                dp[i, j] = -1;
        Console.Write(numberOfPaths(3, 3));
    }
}
 
// This code is contributed by ukasp.

Javascript

        // JavaScript Program of the above approach
 
  var dp = new Array(1001);
  // Loop to create 2D array using 1D array
    for (var i = 0; i < dp.length; i++) {
        dp[i] = new Array(1001);
    }
 
  // Returns count of possible paths to
  // reach cell at row number M and column
  // number N from the topmost leftmost
  // cell (cell at 1, 1)
  function numberOfPaths(M, N)
  {
 
    // If either given row number or
    // given column number is first
    if (M == 1 || N == 1)
      return 1;
 
    // If a value already present
    // in t[][], return it
    if (dp[M][N] != -1) {
      return dp[M][N];
    }
 
    // Horizontal Paths +
    // Vertical Paths +
    // Diagonal Paths
    dp[M][N] = numberOfPaths(M - 1, N)
      + numberOfPaths(M, N - 1)
      + numberOfPaths(M - 1, N - 1);
    return dp[M][N];
  }
 
        // Driver Code
         
      for (let i = 0; i < 1001; i++){
      for (let j = 0; j < 1001; j++){
        dp[i][j] = -1;
      }}
      document.write(numberOfPaths(3, 3));
 
// This code is contributed by avijitmondal1998
    

输出

13

时间复杂度：O(M*N)

辅助空间：O(M*N)