给定大小为n和m的两个数组。我们的任务是找到将给定数组合并为一个数组的方式，以使每个数组的元素顺序不变。
例子：
Input : n = 2, m = 2
Output : 6
Let first array of size n = 2 be [1, 2] and second array of size m = 2 be [3, 4].
So, possible merged array of n + m elements can be:
[1, 2, 3, 4]
[1, 3, 2, 4]
[3, 4, 1, 2]
[3, 1, 4, 2]
[1, 3, 4, 2]
[3, 1, 2, 4]
Input : n = 4, m = 6
Output : 210
这个想法是使用组合学的概念。假设我们有两个数组A {a1，a2，…。，am}和B {b1，b2，…。，bn}分别具有m和n个元素，现在我们必须合并它们而不丢失其顺序。
合并后，我们知道合并后元素的总数将为(m + n)个元素。因此，现在我们只需要从(m + n)中选择m个位置的方法，即可将数组A的元素按其实际顺序放置，即m + n C n 。
放置数组A的m个元素后，将保留n个空格，可以按其实际顺序由B数组的n个元素填充。
因此，合并两个数组以使它们在合并数组中的顺序相同的方式总数为m + n C n
以下是此方法的实现：
C++
// CPP Program to find number of ways
// to merge two array such that their
// order in merged array is same
#include
using namespace std;
// function to find the binomial coefficient
int binomialCoeff(int n, int k)
{
int C[k + 1];
memset(C, 0, sizeof(C));
C[0] = 1; // nC0 is 1
for (int i = 1; i <= n; i++) {
// Compute next row of pascal triangle
// using the previous row
for (int j = min(i, k); j > 0; j--)
C[j] = C[j] + C[j - 1];
}
return C[k];
}
// function to find number of ways
// to merge two array such that their
// order in merged array is same
int numOfWays(int n, int m)
{
return binomialCoeff(m + n, m);
}
// Driven Program
int main()
{
int n = 2, m = 2;
cout << numOfWays(n, m) << endl;
return 0;
} Java
// Java Program to find number of ways
// to merge two array such that their
// order in merged array is same
import java.io.*;
class GFG {
// function to find the binomial
// coefficient
static int binomialCoeff(int n, int k)
{
int C[] = new int[k + 1];
// memset(C, 0, sizeof(C));
C[0] = 1; // nC0 is 1
for (int i = 1; i <= n; i++) {
// Compute next row of pascal
// triangle using the previous
// row
for (int j = Math.min(i, k);
j > 0; j--)
C[j] = C[j] + C[j - 1];
}
return C[k];
}
// function to find number of ways
// to merge two array such that their
// order in merged array is same
static int numOfWays(int n, int m)
{
return binomialCoeff(m + n, m);
}
// Driven Program
public static void main (String[] args)
{
int n = 2, m = 2;
System.out.println(numOfWays(n, m));
}
}
// This code is contributed by anuj_67.Python3
# Python 3 Program to find number of ways
# to merge two array such that their
# order in merged array is same
# function to find the binomial coefficient
def binomialCoeff(n, k):
C = [0 for i in range(k + 1)]
C[0] = 1
for i in range(1, n + 1, 1):
# Compute next row of pascal
# triangle using the previous row
j = min(i, k)
while(j > 0):
C[j] = C[j] + C[j - 1]
j -= 1
return C[k]
# function to find number of ways
# to merge two array such that their
# order in merged array is same
def numOfWays(n, m):
return binomialCoeff(m + n, m)
# Driver Code
if __name__ == '__main__':
n = 2
m = 2
print(numOfWays(n, m))
# This code is contributed by
# Sahil_shelangiaC#
// C# Program to find number of ways
// to merge two array such that their
// order in merged array is same
using System;
class GFG {
// function to find the binomial
// coefficient
static int binomialCoeff(int n, int k)
{
int []C = new int[k + 1];
// memset(C, 0, sizeof(C));
C[0] = 1; // nC0 is 1
for (int i = 1; i <= n; i++) {
// Compute next row of pascal
// triangle using the previous
// row
for (int j = Math.Min(i, k);
j > 0; j--)
C[j] = C[j] + C[j - 1];
}
return C[k];
}
// function to find number of ways
// to merge two array such that their
// order in merged array is same
static int numOfWays(int n, int m)
{
return binomialCoeff(m + n, m);
}
// Driven Program
public static void Main ()
{
int n = 2, m = 2;
Console.WriteLine(numOfWays(n, m));
}
}
// This code is contributed by anuj_67.PHP
0; $j--)
$C[$j] = $C[$j] + $C[$j - 1 ];
}
return $C[$k];
}
// function to find number of ways
// to merge two array such that their
// order in merged array is same
function numOfWays( $n, $m)
{
return binomialCoeff($m + $n, $m);
}
$n = 2; $m = 2;
echo numOfWays($n, $m);
//This code is contributed by Rajput-Ji.
?>输出：
6
我们可以使用二项式系数的线性时间实现来解决上述线性时间问题。