计算距离等于曼哈顿距离的路径

给定二维坐标系中的两个点(x1, y1)和(x2, y2) 。任务是计算距离等于两个给定点之间的曼哈顿距离的所有路径。
例子：

Input: x1 = 2, y1 = 3, x2 = 4, y2 = 5
Output: 6
Input: x1 = 2, y1 = 6, x2 = 4, y2 = 9
Output: 10

方法：点(x1, y1)和(x2, y2)之间的曼哈顿距离将为abs(x1 – x2) + abs(y1 – y2)
让abs(x1 – x2) = m和abs(y1 – y2) = n
每条距离等于曼哈顿距离的路径总是有m+n 条边， m 条水平边和n 条垂直边。因此，这是基于组形成的组合数学的基本情况。这背后的想法是将(m + n)种不同事物分为两组的方式数量，一组包含m 个项目，另一个包含n 个项目，由^{m + n} C _n给出，即(m + n) ！ /米！ * n！ .

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define ll long long int
 
// Function to return the value of nCk
ll binomialCoeff(int n, int k)
{
    ll res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n * (n-1) *---* (n-k+1)] /
    // [k * (k-1) *---* 1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// Function to return the number of paths
ll countPaths(int x1, int y1, int x2, int y2)
{
 
    // Difference between the 'x'
    // coordinates of the given points
    int m = abs(x1 - x2);
 
    // Difference between the 'y'
    // coordinates of the given points
    int n = abs(y1 - y2);
 
    return (binomialCoeff(m + n, n));
}
 
// Driver code
int main()
{
    int x1 = 2, y1 = 3, x2 = 4, y2 = 5;
    cout << countPaths(x1, y1, x2, y2);
    return 0;
}

Java

// Java implementation of the approach
class GfG
{
     
//static long ll long long int
 
// Function to return the value of nCk
static long binomialCoeff(int n, int k)
{
    long res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n * (n-1) *---* (n-k+1)] /
    // [k * (k-1) *---* 1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// Function to return the number of paths
static long countPaths(int x1, int y1, int x2, int y2)
{
 
    // Difference between the 'x'
    // coordinates of the given points
    int m = Math.abs(x1 - x2);
 
    // Difference between the 'y'
    // coordinates of the given points
    int n = Math.abs(y1 - y2);
 
    return (binomialCoeff(m + n, n));
}
 
// Driver code
public static void main(String[] args)
{
    int x1 = 2, y1 = 3, x2 = 4, y2 = 5;
    System.out.println(countPaths(x1, y1, x2, y2));
}
}
 
// This code is contributed by
// Prerna Saini.

Python3

# Python3 implementation of the approach
 
# Function to return the value of nCk
def binomialCoeff(n, k):
 
    res = 1
 
    # Since C(n, k) = C(n, n-k)
    if (k > n - k):
        k = n - k
 
    # Calculate value of
    # [n * (n-1) *---* (n-k+1)] /
    # [k * (k-1) *---* 1]
    for i in range(k):
        res *= (n - i)
        res //= (i + 1)
 
    return res
 
# Function to return the number
# of paths
def countPaths(x1, y1, x2, y2):
 
    # Difference between the 'x'
    # coordinates of the given points
    m = abs(x1 - x2)
 
    # Difference between the 'y'
    # coordinates of the given points
    n = abs(y1 - y2)
 
    return (binomialCoeff(m + n, n))
 
# Driver code
x1, y1, x2, y2 = 2, 3, 4, 5
print(countPaths(x1, y1, x2, y2))
 
# This code is contributed
# by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the value of nCk
static long binomialCoeff(int n, int k)
{
    long res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n * (n-1) *---* (n-k+1)] /
    // [k * (k-1) *---* 1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// Function to return the number of paths
static long countPaths(int x1, int y1,
                       int x2, int y2)
{
 
    // Difference between the 'x'
    // coordinates of the given points
    int m = Math.Abs(x1 - x2);
 
    // Difference between the 'y'
    // coordinates of the given points
    int n = Math.Abs(y1 - y2);
 
    return (binomialCoeff(m + n, n));
}
 
// Driver code
public static void Main()
{
    int x1 = 2, y1 = 3, x2 = 4, y2 = 5;
    Console.Write(countPaths(x1, y1, x2, y2));
}
}
 
// This code is contributed by
// Akanksha Rai

PHP

 $n - $k)
        $k = $n - $k;
 
    // Calculate value of
    // [n * (n-1) *---* (n-k+1)] /
    // [k * (k-1) *---* 1]
    for ($i = 0; $i < $k; ++$i)
    {
        $res *= ($n - $i);
        $res /= ($i + 1);
    }
 
    return $res;
}
 
// Function to return the number of paths
function countPaths($x1, $y1, $x2, $y2)
{
 
    // Difference between the 'x'
    // coordinates of the given points
    $m =abs($x1 - $x2);
 
    // Difference between the 'y'
    // coordinates of the given points
    $n = abs($y1 - $y2);
 
    return (binomialCoeff($m + $n, $n));
}
 
// Driver code
{
    $x1 = 2; $y1 = 3; $x2 = 4; $y2 = 5;
    echo(countPaths($x1, $y1, $x2, $y2));
}
 
// This code is contributed by
// Code_Mech.

Javascript

输出：