计算 mXn 矩阵从左上角到右下角的所有可能路径

问题是计算 mXn 矩阵从左上角到右下角的所有可能路径，其约束条件是从每个单元格只能向右或向下移动
例子 ：

Input :  m = 2, n = 2;
Output : 2
There are two paths
(0, 0) -> (0, 1) -> (1, 1)
(0, 0) -> (1, 0) -> (1, 1)

Input :  m = 2, n = 3;
Output : 3
There are three paths
(0, 0) -> (0, 1) -> (0, 2) -> (1, 2)
(0, 0) -> (0, 1) -> (1, 1) -> (1, 2)
(0, 0) -> (1, 0) -> (1, 1) -> (1, 2)

我们已经讨论了打印所有可能路径的解决方案，计算所有路径更容易。令 NumberOfPaths(m, n) 为到达矩阵中行号 m 和列号 n 的路径数，NumberOfPaths(m, n) 可以递归地写成如下。

C++

// A C++  program to count all possible paths
// from top left to bottom right
 
#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 is first or given column
    // number is first
    if (m == 1 || n == 1)
        return 1;
 
    // If diagonal movements are allowed then the last
    // addition is required.
    return numberOfPaths(m - 1, n) + numberOfPaths(m, n - 1);
    // + numberOfPaths(m-1, n-1);
}
 
int main()
{
    cout << numberOfPaths(3, 3);
    return 0;
}

Java

// A Java program to count all possible paths
// from top left to bottom right
 
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 is first or
        // given column number is first
        if (m == 1 || n == 1)
            return 1;
 
        // If diagonal movements are allowed then
        // the last addition is required.
        return numberOfPaths(m - 1, n) + numberOfPaths(m, n - 1);
        // + numberOfPaths(m-1, n-1);
    }
 
    public static void main(String args[])
    {
        System.out.println(numberOfPaths(3, 3));
    }
}
 
// This code is contributed by Sumit Ghosh

Python

# Python program to count all possible paths
# from top left to bottom right
 
# function to return 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 is first
# or given column number is first
    if(m == 1 or n == 1):
        return 1
 
# If diagonal movements are allowed
# then the last addition
# is required.
    return numberOfPaths(m-1, n) + numberOfPaths(m, n-1)
 
# Driver program to test above function
m = 3
n = 3
print(numberOfPaths(m, n))
 
# This code is contributed by Aditi Sharma

C#

// A C# program to count all possible paths
// from top left to bottom right
 
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 is first or
        // given column number is first
        if (m == 1 || n == 1)
            return 1;
 
        // If diagonal movements are allowed then
        // the last addition is required.
        return numberOfPaths(m - 1, n) + numberOfPaths(m, n - 1);
        // + numberOfPaths(m-1, n-1);
    }
 
    static public void Main()
    {
        Console.WriteLine(numberOfPaths(3, 3));
    }
}
 
// This code is contributed by ajit

Javascript

// A Javascript program to count all possible paths
// from top left to bottom right
     
    // 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 is first or
        // given column number is first
        if (m == 1 || n == 1)
            return 1;
         
        // If diagonal movements are allowed then
        // the last addition is required.
        return numberOfPaths(m - 1, n) + numberOfPaths(m, n - 1);
        
       // + numberOfPaths(m - 1, n - 1);
    }
     
    document.write(numberOfPaths(3, 3)+"<br>");
     
    // This code is contributed by rag2127

---

输出：

6

上述递归解决方案的时间复杂度是指数级的。有许多重叠的子问题。我们可以为 numberOfPaths(3, 3) 绘制一个递归树，并看到许多重叠的子问题。递归树类似于最长公共子序列问题的递归树。
所以这个问题具有动态规划问题的两个性质(见this和this)。与其他典型的动态规划 (DP) 问题一样，通过使用上述递归公式以自下而上的方式构造临时数组 count[][] 可以避免相同子问题的重新计算。

C++

// A C++ program to count all possible paths
// from top left to bottom right
#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)
{
    // Create a 2D table to store results of subproblems
    int count[m][n];
 
    // Count of paths to reach any cell in first column is 1
    for (int i = 0; i < m; i++)
        count[i][0] = 1;
 
    // Count of paths to reach any cell in first row is 1
    for (int j = 0; j < n; j++)
        count[0][j] = 1;
 
    // Calculate count of paths for other cells in
    // bottom-up manner using the recursive solution
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++)
 
            // By uncommenting the last part the code calculates the total
            // possible paths if the diagonal Movements are allowed
            count[i][j] = count[i - 1][j] + count[i][j - 1]; //+ count[i-1][j-1];
    }
    return count[m - 1][n - 1];
}
 
// Driver program to test above functions
int main()
{
    cout << numberOfPaths(3, 3);
    return 0;
}

Java

// A Java program to count all possible paths
// from top left to bottom right
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)
    {
        // Create a 2D table to store results
        // of subproblems
        int count[][] = new int[m][n];
 
        // Count of paths to reach any cell in
        // first column is 1
        for (int i = 0; i < m; i++)
            count[i][0] = 1;
 
        // Count of paths to reach any cell in
        // first column is 1
        for (int j = 0; j < n; j++)
            count[0][j] = 1;
 
        // Calculate count of paths for other
        // cells in bottom-up manner using
        // the recursive solution
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++)
 
                // By uncommenting the last part the
                // code calculates the total possible paths
                // if the diagonal Movements are allowed
                count[i][j] = count[i - 1][j] + count[i][j - 1]; //+ count[i-1][j-1];
        }
        return count[m - 1][n - 1];
    }
 
    // Driver program to test above function
    public static void main(String args[])
    {
        System.out.println(numberOfPaths(3, 3));
    }
}
 
// This code is contributed by Sumit Ghosh

Python

# Python program to count all possible paths
# from top left to bottom right
 
# 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):
    # Create a 2D table to store
    # results of subproblems
    # one-liner logic to take input for rows and columns
    # mat = [[int(input()) for x in range (C)] for y in range(R)]
     
    count = [[0 for x in range(n)] for y in range(m)]
   
    # Count of paths to reach any
    # cell in first column is 1
    for i in range(m):
        count[i][0] = 1;
   
    # Count of paths to reach any
    # cell in first column is 1
    for j in range(n):
        count[0][j] = 1;
   
    # Calculate count of paths for other
    # cells in bottom-up
    # manner using the recursive solution
    for i in range(1, m):
        for j in range(1, n):            
            count[i][j] = count[i-1][j] + count[i][j-1]
    return count[m-1][n-1]
 
# Driver program to test above function
m = 3
n = 3
print( numberOfPaths(m, n))
 
# This code is contributed by Aditi Sharma

C＃

// A C#  program to count all possible paths
// from top left to bottom right
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)
    {
        // Create a 2D table to store results
        // of subproblems
        int[, ] count = new int[m, n];
 
        // Count of paths to reach any cell in
        // first column is 1
        for (int i = 0; i < m; i++)
            count[i, 0] = 1;
 
        // Count of paths to reach any cell in
        // first column is 1
        for (int j = 0; j < n; j++)
            count[0, j] = 1;
 
        // Calculate count of paths for other
        // cells in bottom-up manner using
        // the recursive solution
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++)
 
                // By uncommenting the last part the
                // code calculates the total possible paths
                // if the diagonal Movements are allowed
                count[i, j] = count[i - 1, j] + count[i, j - 1]; //+ count[i-1][j-1];
        }
        return count[m - 1, n - 1];
    }
 
    // Driver program to test above function
    static public void Main()
    {
        Console.WriteLine(numberOfPaths(3, 3));
    }
}
 
// This code is contributed by akt_mit

Javascript

 
// A javascript program to count all possible paths
// from top left to bottom right
 
// 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)
{
    // Create a 2D table to store results
    // of subproblems
    var count = Array(m).fill(0).map(x => Array(n).fill(0));
 
    // Count of paths to reach any cell in
    // first column is 1
    for (i = 0; i < m; i++)
        count[i][0] = 1;
 
    // Count of paths to reach any cell in
    // first column is 1
    for (j = 0; j < n; j++)
        count[0][j] = 1;
 
    // Calculate count of paths for other
    // cells in bottom-up manner using
    // the recursive solution
    for (i = 1; i < m; i++) {
        for (j = 1; j < n; j++)
 
            // By uncommenting the last part the
            // code calculates the total possible paths
            // if the diagonal Movements are allowed
             
            //+ count[i-1][j-1];
            count[i][j] = count[i - 1][j] + count[i][j - 1];
    }
    return count[m - 1][n - 1];
}
 
// Driver program to test above function
document.write(numberOfPaths(3, 3));
 
// This code is contributed by 29AjayKumar
 

输出：

6

上述动态规划解决方案的时间复杂度为 O(mn)。
上述解的空间复杂度为O(mn)。
DP 解决方案的空间优化。
上面的解决方案更直观，但我们也可以将空间减少 O(n)；其中 n 是列大小。

C++

#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)
{
    // Create a 1D array to store results of subproblems
    int dp[n] = { 1 };
    dp[0] = 1;
 
    for (int i = 0; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[j] += dp[j - 1];
        }
    }
 
    return dp[n - 1];
}
 
// Driver code
int main()
{
    cout << numberOfPaths(3, 3);
}
 
// This code is contributed by mohit kumar 29

Java

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)
    {
        // Create a 1D array to store results of subproblems
        int[] dp = new int[n];
        dp[0] = 1;
 
        for (int i = 0; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[j] += dp[j - 1];
            }
        }
 
        return dp[n - 1];
    }
 
    // Driver program to test above function
    public static void main(String args[])
    {
        System.out.println(numberOfPaths(3, 3));
    }
}

蟒蛇3

# 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(p, q):
     
    # Create a 1D array to store
    # results of subproblems
    dp = [1 for i in range(q)]
    for i in range(p - 1):
        for j in range(1, q):
            dp[j] += dp[j - 1]
    return dp[q - 1]
 
# Driver Code
print(numberOfPaths(3, 3))
 
# This code is contributed
# by Ankit Yadav

C＃

using System;
 
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)
    {
        // Create a 1D array to store
        // results of subproblems
        int[] dp = new int[n];
        dp[0] = 1;
 
        for (int i = 0; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[j] += dp[j - 1];
            }
        }
 
        return dp[n - 1];
    }
 
    // Driver Code
    public static void Main()
    {
        Console.Write(numberOfPaths(3, 3));
    }
}
 
// This code is contributed
// by ChitraNayal

Javascript

 
// 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)
{
    // Create a 1D array to store results of subproblems
    dp = Array.from({length: n}, (_, i) => 0);
    dp[0] = 1;
 
    for (i = 0; i < m; i++) {
        for (j = 1; j < n; j++) {
            dp[j] += dp[j - 1];
        }
    }
 
    return dp[n - 1];
}
 
// Driver program to test above function
document.write(numberOfPaths(3, 3));
 
// This code contributed by Princi Singh
 

输出：

6

此代码由Vivek Singh 提供
请注意，也可以使用公式 (m-1 + n-1)!/(m-1)!(n-1)! 计算计数。
另一种方法：(使用组合学)在这种方法中，我们必须在这里计算 m+n-2 C n-1 ，这将是 (m+n-2)！ / (n-1)！ (米-1)！

C++

// A C++ program to count all possible paths from
// top left to top bottom using combinatorics
 
#include 
using namespace std;
 
int numberOfPaths(int m, int n)
{
    // We have to calculate m+n-2 C n-1 here
    // which will be (m+n-2)! / (n-1)! (m-1)!
    int path = 1;
    for (int i = n; i < (m + n - 1); i++) {
        path *= i;
        path /= (i - n + 1);
    }
    return path;
}
 
// Driver code
int main()
{
    cout << numberOfPaths(3, 3);
    return 0;
}
 
// This code is suggested by Kartik Sapra

Java

// Java program to count all possible paths from
// top left to top bottom using combinatorics
class GFG {
 
    static int numberOfPaths(int m, int n)
    {
        // We have to calculate m+n-2 C n-1 here
        // which will be (m+n-2)! / (n-1)! (m-1)!
        int path = 1;
        for (int i = n; i < (m + n - 1); i++) {
            path *= i;
            path /= (i - n + 1);
        }
        return path;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        System.out.println(numberOfPaths(3, 3));
    }
}
 
// This code is contributed by Code_Mech.

蟒蛇3

# Python3 program to count all possible
# paths from top left to top bottom
# using combinatorics
def numberOfPaths(m, n) :
 
    # We have to calculate m + n-2 C n-1 here
    # which will be (m + n-2)! / (n-1)! (m-1)! path = 1;
    for i in range(n, (m + n - 1)):
        path *= i;
        path //= (i - n + 1);
     
    return path;
 
# Driver code
print(numberOfPaths(3, 3));
 
# This code is contributed
# by Akanksha Rai

C＃

// C# program to count all possible paths from
// top left to top bottom using combinatorics
using System;
 
class GFG {
 
    static int numberOfPaths(int m, int n)
    {
        // We have to calculate m+n-2 C n-1 here
        // which will be (m+n-2)! / (n-1)! (m-1)!
        int path = 1;
        for (int i = n; i < (m + n - 1); i++) {
            path *= i;
            path /= (i - n + 1);
        }
        return path;
    }
 
    // Driver code
    public static void Main()
    {
        Console.WriteLine(numberOfPaths(3, 3));
    }
}
 
// This code is contributed by Code_Mech.

Javascript

 
// javascript program to count all possible paths from
// top left to top bottom using combinatorics  
function numberOfPaths(m , n)
    {
        // We have to calculate m+n-2 C n-1 here
        // which will be (m+n-2)! / (n-1)! (m-1)!
        var path = 1;
        for (i = n; i < (m + n - 1); i++) {
            path *= i;
            path = parseInt(path/(i - n + 1));
        }
        return path;
    }
 
// Driver code
document.write(numberOfPaths(3, 3));
 
// This code contributed by Princi Singh

输出：

6

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