给定一个数字n，创建一个大小为2n的数组，以使该数组包含从1到n的每个数字的2个实例，并且在数量为i的两个实例之间的元素数等于i。如果无法进行这种配置，请打印相同的内容。
例子：
Input: n = 3
Output: res[] = {3, 1, 2, 1, 3, 2}
Input: n = 2
Output: Not Possible
Input: n = 4
Output: res[] = {4, 1, 3, 1, 2, 4, 3, 2}强烈建议您最小化浏览器，然后自己尝试。
一种解决方案是回溯。这个想法很简单，我们将n的两个实例放在一个位置，然后递归n-1。如果重复成功，则返回true，否则我们回溯并尝试将n放置在其他位置。以下是该想法的实现。
C++
// A backtracking based C++ Program to fill
// two instances of all numbers from 1 to n
// in a specific way
#include
using namespace std;
// A recursive utility function to fill
// two instances of numbers from 1 to n
// in res[0..2n-1]. 'curr' is current value of n.
bool fillUtil(int res[], int curr, int n)
{
// If current number becomes 0,
// then all numbers are filled
if (curr == 0)
return true;
// Try placing two instances of 'curr' at
// all possible locations till solution is found
int i;
for (i = 0; i < 2 * n - curr - 1; i++)
{
// Two 'curr' should be placed at
// 'curr+1' distance
if (res[i] == 0 && res[i + curr + 1] == 0)
{
// Plave two instances of 'curr'
res[i] = res[i + curr + 1] = curr;
// Recur to check if the above placement
// leads to a solution
if (fillUtil(res, curr - 1, n))
return true;
// If solution is not possible,
// then backtrack
res[i] = res[i + curr + 1] = 0;
}
}
return false;
}
// This function prints the result for
// input number 'n' using fillUtil()
void fill(int n)
{
// Create an array of size 2n and
// initialize all elements in it as 0
int res[2 * n], i;
for (i = 0; i < 2 * n; i++)
res[i] = 0;
// If solution is possible,
// then print it.
if (fillUtil(res, n, n))
{
for (i = 0; i < 2 * n; i++)
cout << res[i] << " ";
}
else
cout << "Not Possible";
}
// Driver Code
int main()
{
fill(7);
return 0;
}
// This code is contributed
// by SHUBHAMSINGH8410 C
// A backtracking based C Program to fill two instances of all numbers
// from 1 to n in a specific way
#include
#include
// A recursive utility function to fill two instances of numbers from
// 1 to n in res[0..2n-1]. 'curr' is current value of n.
bool fillUtil(int res[], int curr, int n)
{
// If current number becomes 0, then all numbers are filled
if (curr == 0) return true;
// Try placing two instances of 'curr' at all possible locations
// till solution is found
int i;
for (i=0; i<2*n-curr-1; i++)
{
// Two 'curr' should be placed at 'curr+1' distance
if (res[i] == 0 && res[i + curr + 1] == 0)
{
// Plave two instances of 'curr'
res[i] = res[i + curr + 1] = curr;
// Recur to check if the above placement leads to a solution
if (fillUtil(res, curr-1, n))
return true;
// If solution is not possible, then backtrack
res[i] = res[i + curr + 1] = 0;
}
}
return false;
}
// This function prints the result for input number 'n' using fillUtil()
void fill(int n)
{
// Create an array of size 2n and initialize all elements in it as 0
int res[2*n], i;
for (i=0; i<2*n; i++)
res[i] = 0;
// If solution is possible, then print it.
if (fillUtil(res, n, n))
{
for (i=0; i<2*n; i++)
printf("%d ", res[i]);
}
else
puts("Not Possible");
}
// Driver program
int main()
{
fill(7);
return 0;
} Java
// A backtracking based C++ Program to fill
// two instances of all numbers from 1 to n
// in a specific way
import java.io.*;
class GFG
{
// A recursive utility function to fill
// two instances of numbers from 1 to n
// in res[0..2n-1]. 'curr' is current value of n.
static boolean fillUtil(int res[], int curr, int n)
{
// If current number becomes 0,
// then all numbers are filled
if (curr == 0)
return true;
// Try placing two instances of 'curr' at
// all possible locations till solution is found
int i;
for (i = 0; i < 2 * n - curr - 1; i++)
{
// Two 'curr' should be placed at
// 'curr+1' distance
if (res[i] == 0 && res[i + curr + 1] == 0)
{
// Plave two instances of 'curr'
res[i] = res[i + curr + 1] = curr;
// Recur to check if the above placement
// leads to a solution
if (fillUtil(res, curr - 1, n))
return true;
// If solution is not possible,
// then backtrack
res[i] = res[i + curr + 1] = 0;
}
}
return false;
}
// This function prints the result for
// input number 'n' using fillUtil()
static void fill(int n)
{
// Create an array of size 2n and
// initialize all elements in it as 0
int res[] = new int[2 * n];
int i;
for (i = 0; i < 2 * n; i++)
res[i] = 0;
// If solution is possible,
// then print it.
if (fillUtil(res, n, n))
{
for (i = 0; i < 2 * n; i++)
System.out.print(res[i] + " ");
}
else
System.out.print("Not Possible");
}
// Driver Code
public static void main (String[] args)
{
fill(7);
}
}
// This code is contributed by ajitPython3
# A backtracking based Python3 Program
# to fill two instances of all numbers
# from 1 to n in a specific way
def fillUtil(res, curr, n):
# A recursive utility function to fill
# two instances of numbers from 1 to n
# in res[0..2n-1]. 'curr' is current value of n.
# If current number becomes 0,
# then all numbers are filled
if curr == 0:
return True
# Try placing two instances of 'curr' at all
# possible locations till solution is found
for i in range(2 * n - curr - 1):
# Two 'curr' should be placed
# at 'curr+1' distance
if res[i] == 0 and res[i + curr + 1] == 0:
# Place two instances of 'curr'
res[i] = res[i + curr + 1] = curr
# Recur to check if the above
# placement leads to a solution
if fillUtil(res, curr - 1, n):
return True
# If solution is not possible,
# then backtrack
res[i] = 0
res[i + curr + 1] = 0
return False
def fill(n):
# This function prints the result
# for input number 'n' using fillUtil()
# Create an array of size 2n and
# initialize all elements in it as 0
res = [0] * (2 * n)
# If solution is possible, then print it.
if fillUtil(res, n, n):
for i in range(2 * n):
print(res[i], end = ' ')
print()
else:
print("Not Possible")
# Driver Code
if __name__ == '__main__':
fill(7)
# This code is contributed by vibhu4agarwalC#
// A backtracking based C# Program to fill
// two instances of all numbers from 1 to n
// in a specific way
using System;
class GFG
{
// A recursive utility function to fill
// two instances of numbers from 1 to n
// in res[0..2n-1]. 'curr' is current value of n.
static bool fillUtil(int []res, int curr, int n)
{
// If current number becomes 0,
// then all numbers are filled
if (curr == 0)
return true;
// Try placing two instances of 'curr' at
// all possible locations till solution is found
int i;
for (i = 0; i < 2 * n - curr - 1; i++)
{
// Two 'curr' should be placed at
// 'curr+1' distance
if (res[i] == 0 && res[i + curr + 1] == 0)
{
// Plave two instances of 'curr'
res[i] = res[i + curr + 1] = curr;
// Recur to check if the above placement
// leads to a solution
if (fillUtil(res, curr - 1, n))
return true;
// If solution is not possible,
// then backtrack
res[i] = res[i + curr + 1] = 0;
}
}
return false;
}
// This function prints the result for
// input number 'n' using fillUtil()
static void fill(int n)
{
// Create an array of size 2n and
// initialize all elements in it as 0
int []res=new int[2 * n];
int i;
for (i = 0; i < (2 * n); i++)
res[i] = 0;
// If solution is possible,
// then print it.
if (fillUtil(res, n, n))
{
for (i = 0; i < 2 * n; i++)
Console.Write (res[i] + " ");
}
else
Console.Write ("Not Possible");
}
// Driver Code
static public void Main ()
{
fill(7);
}
}
// This code is contributed by ajitJavascript
输出：
7 3 6 2 5 3 2 4 7 6 5 1 4 1