从网格中左下角的单元格到达每个单元格的最短路径数

给定两个数N和M 。任务是在大小为N × M的网格中找到从左下角开始移动时到达单元格 (i, j)的最短路径数

注： cell(i, j) 代表网格中的第 i 行第 j 列

下图显示了在4 × 4网格中到达cell(1, 4) 的一些最短路径

例子 ：

Input : N = 3, M = 4 
Output : 1 3 6 10 
         1 2 3 4 
         1 1 1 1  

Input : N = 5, M = 2 
Output : 1 5 
         1 4 
         1 3 
         1 2 
         1 1 

方法：一种有效的方法是从左下角开始计算网格。

• 到达 cell(n, i) 的最短路径数为 1，其中 1 < = i < = M
• 到达 cell(i, 1) 的最短路径数为 1，其中 1 < = i < = N
• 到达cell(i, j)的最短路径数是cell(i-1, j)和(i, j+1)的最短路径数之和，其中，1 < = j < = M and 1 < = i < = N

下面是上述方法的实现：

C++

// CPP program to find number of shortest paths
#include 
using namespace std;
  
// Function to find number of shortest paths
void NumberOfShortestPaths(int n, int m)
{
    int a[n][m];
  
    for (int i = 0; i < n; i++)
        memset(a[i], 0, sizeof(a[i]));
  
    // Compute the grid starting from
    // the bottom-left corner
    for (int i = n - 1; i >= 0; i--) {
        for (int j = 0; j < m; j++) {
            if (j == 0 or i == n - 1)
                a[i][j] = 1;
            else
                a[i][j] = a[i][j - 1] + a[i + 1][j];
        }
    }
  
    // Print the grid
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cout << a[i][j] << " ";
        }
        cout << endl;
    }
}
  
// Driver code
int main()
{
    int n = 5, m = 2;
  
    // Function call
    NumberOfShortestPaths(n, m);
  
    return 0;
}

Java

// Java program to find number of shortest paths
class GFG
{
  
// Function to find number of shortest paths
static void NumberOfShortestPaths(int n, int m)
{
    int [][]a = new int[n][m];
  
    // Compute the grid starting from
    // the bottom-left corner
    for (int i = n - 1; i >= 0; i--) 
    {
        for (int j = 0; j < m; j++) 
        {
            if (j == 0 || i == n - 1)
                a[i][j] = 1;
            else
                a[i][j] = a[i][j - 1] + a[i + 1][j];
        }
    }
  
    // Print the grid
    for (int i = 0; i < n; i++) 
    {
        for (int j = 0; j < m; j++) 
        {
            System.out.print(a[i][j] + " ");
        }
        System.out.println();
    }
}
  
// Driver code
public static void main(String[] args)
{
    int n = 5, m = 2;
  
    // Function call
    NumberOfShortestPaths(n, m);
}
}
  
// This code is contributed by Princi Singh

Python3

# Python 3 program to find 
# number of shortest paths
  
# Function to find number of shortest paths
def NumberOfShortestPaths(n, m):
    a = [[0 for i in range(m)]
            for j in range(n)]
  
    for i in range(n):
        for j in range(m):
            a[i][j] = 0
  
    # Compute the grid starting from
    # the bottom-left corner
    i = n - 1
    while(i >= 0):
        for j in range(m):
            if (j == 0 or i == n - 1):
                a[i][j] = 1
            else:
                a[i][j] = a[i][j - 1] + \
                          a[i + 1][j]
  
        i -= 1
  
    # Print the grid
    for i in range(n):
        for j in range(m):
            print(a[i][j], end = " ")
        print("\n", end = "")
  
# Driver code
if __name__ == '__main__':
    n = 5
    m = 2
  
    # Function call
    NumberOfShortestPaths(n, m)
      
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find number of shortest paths
using System;
  
class GFG
{
  
// Function to find number of shortest paths
static void NumberOfShortestPaths(int n, int m)
{
    int [,]a = new int[n, m];
  
    // Compute the grid starting from
    // the bottom-left corner
    for (int i = n - 1; i >= 0; i--) 
    {
        for (int j = 0; j < m; j++) 
        {
            if (j == 0 || i == n - 1)
                a[i, j] = 1;
            else
                a[i, j] = a[i, j - 1] + a[i + 1, j];
        }
    }
  
    // Print the grid
    for (int i = 0; i < n; i++) 
    {
        for (int j = 0; j < m; j++) 
        {
            Console.Write(a[i, j] + " ");
        }
        Console.Write("\n");
    }
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 5, m = 2;
  
    // Function call
    NumberOfShortestPaths(n, m);
}
}
  
// This code is contributed by PrinciRaj1992

输出 ：

1 5 
1 4 
1 3 
1 2 
1 1 

时间复杂度： O(N × M)

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