给定两个整数N和K ，其中K
例子：
Input: N = 3, K = 2
Output: 1 3 2
|1 – 3| = 2 and |3 – 2| = 1 which gives 2 distinct integers (2 and 1)
Input: N = 5, K = 4
Output: 1 5 2 4 3
|1 – 5| = 4, |5 – 2| = 3, |2 – 4| = 2 and |4 – 3| = 1 gives 4 distinct integers i.e. 4, 3, 2 and 1
方法：通过简单的观察就可以轻松解决问题。在奇数索引处，递增顺序为1、2、3，… ，在偶数索引处，递减顺序为N，N-1，N-2，… ，依此类推。
对于N = 10 ，对于连续的绝对差，具有不同整数的置换可以是1 10 2 9 3 8 4 7 5 6 。连续的绝对差给出整数9、8、7 ，依此类推。
因此，首先打印此序列的K个整数，然后使其余的差等于1 。该代码很容易说明。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to generate a permutation of integers
// from 1 to N such that the absolute difference of
// all the two consecutive integers give K distinct integers
void printPermutation(int N, int K)
{
// To store the permutation
vector res;
int l = 1, r = N, flag = 0;
for (int i = 0; i < K; i++) {
if (!flag) {
// For sequence 1 2 3...
res.push_back(l);
l++;
}
else {
// For sequence N, N-1, N-2...
res.push_back(r);
r--;
}
// Flag is used to alternate between
// the above if else statements
flag ^= 1;
}
// Taking integers with difference 1
// If last element added was r + 1
if (!flag) {
for (int i = r; i >= l; i--)
res.push_back(i);
}
// If last element added was l - 1
else {
for (int i = l; i <= r; i++)
res.push_back(i);
}
// Print the permutation
for (auto i : res)
cout << i << " ";
}
// Driver code
int main()
{
int N = 10, K = 4;
printPermutation(N, K);
return 0;
} Java
// Java implementation of the above approach
import java.util.Vector;
class GFG
{
// Function to generate a permutation
// of integers from 1 to N such that the
// absolute difference of all the two
// consecutive integers give K distinct integers
static void printPermutation(int N, int K)
{
// To store the permutation
Vector res = new Vector<>();
int l = 1, r = N, flag = 0;
for (int i = 0; i < K; i++)
{
if (flag == 0)
{
// For sequence 1 2 3...
res.add(l);
l++;
}
else
{
// For sequence N, N-1, N-2...
res.add(r);
r--;
}
// Flag is used to alternate between
// the above if else statements
flag ^= 1;
}
// Taking integers with difference 1
// If last element added was r + 1
if (flag != 1)
{
for (int i = r; i >= l; i--)
{
res.add(i);
}
}
// If last element added was l - 1
else
{
for (int i = l; i <= r; i++)
{
res.add(i);
}
}
// Print the permutation
for (Integer i : res)
{
System.out.print(i + " ");
}
}
// Driver code
public static void main(String[] args)
{
int N = 10, K = 4;
printPermutation(N, K);
}
}
// This code is contributed by
// 29AjayKumar Python3
# Python 3 implementation of the approach
# Function to generate a permutation
# of integers from 1 to N such that the
# absolute difference of all the two
# consecutive integers give K distinct
# integers
def printPermutation(N, K):
# To store the permutation
res = list();
l, r, flag = 1, N, 0
for i in range(K):
if flag == False:
# For sequence 1 2 3...
res.append(l)
l += 1
else:
# For sequence N, N-1, N-2...
res.append(r);
r -= 1;
# Flag is used to alternate between
# the above if else statements
flag = flag ^ 1;
# Taking integers with difference 1
# If last element added was r + 1
if flag == False:
for i in range(r, 2, -1):
res.append(i)
# If last element added was l - 1
else:
for i in range(l, r):
res.append(i)
# Print the permutation
for i in res:
print(i, end = " ")
# Driver code
N, K = 10, 4
printPermutation(N, K)
# This code is contributed by
# Mohit Kumar 29C#
// C# implementation of the above approach
using System;
using System.Collections;
class GFG
{
// Function to generate a permutation
// of integers from 1 to N such that the
// absolute difference of all the two
// consecutive integers give K distinct integers
static void printPermutation(int N, int K)
{
// To store the permutation
ArrayList res = new ArrayList();
int l = 1, r = N, flag = 0;
for (int i = 0; i < K; i++)
{
if (flag == 0)
{
// For sequence 1 2 3...
res.Add(l);
l++;
}
else
{
// For sequence N, N-1, N-2...
res.Add(r);
r--;
}
// Flag is used to alternate between
// the above if else statements
flag ^= 1;
}
// Taking integers with difference 1
// If last element added was r + 1
if (flag != 1)
{
for (int i = r; i >= l; i--)
{
res.Add(i);
}
}
// If last element added was l - 1
else
{
for (int i = l; i <= r; i++)
{
res.Add(i);
}
}
// Print the permutation
foreach (int i in res)
{
Console.Write(i + " ");
}
}
// Driver code
public static void Main()
{
int N = 10, K = 4;
printPermutation(N, K);
}
}
// This code is contributed by PrinciRaj1992PHP
= $l; $i--)
array_push($res, $i);
}
// If last element added was l - 1
else
{
for ($i = l; $i <= $r; $i++)
array_push($res, $i);
}
// Print the permutation
for($i = 0; $i < sizeof($res); $i++)
echo $res[$i], " ";
}
// Driver code
$N = 10;
$K = 4;
printPermutation($N, $K);
// This code is contributed by Ryuga
?>输出：
1 10 2 9 8 7 6 5 4 3