K 差排列
给定两个整数 n 和 k。考虑自然数 n 的第一个排列，P =“1 2 3…n”，打印一个排列“结果”，使得 abs(结果I–PI)= k，其中 P i 表示 I 在排列 P 中的位置。P i 的值从 1 到 n 不等。如果有多个可能的结果，则打印字典上最小的一个。
Input: n = 6 k = 3
Output: 4 5 6 1 2 3
Explanation:
P = 1 2 3 4 5 6
Result = 4 5 6 1 2 3
We can notice that the difference between
individual numbers (at same positions) of
P and result is 3 and "4 5 6 1 2 3" is
lexicographically smallest such permutation.
Other greater permutations could be
Input : n = 6 k = 2
Output : Not possible
Explanation: No permutation is possible
with difference is k
朴素方法是生成从 1 到 n 的所有排列，选出满足绝对差 k 条件的最小排列，该方法的时间复杂度为ω(n！)这对于大数值的 n. 肯定会超时有效的方法是观察指数每个位置的模式。对于索引 I 的每个位置，只能存在两个候选，即 i + k 和 I–k。由于我们需要找到字典上最小的排列，因此我们将首先寻找 I–k 候选(如果可能)，然后寻找 i + k 候选。
Illustration:
n = 8, k = 2
P : 1 2 3 4 5 6 7 8
For any i<sup>th position we will check which candidate</sup>
<sup>is possible i.e., i + k or i - k</sup>
1st pos = 1 + 2 = 3 (1 - 2 not possible)
2nd pos = 2 + 2 = 4 (2 - 2 not possible)
3rd pos = 3 - 2 = 1 (possible)
4th pos = 4 - 2 = 2 (possible)
5th pos = 5 + 2 = 7 (5 - 2 already placed, not possible)
6th pos = 6 + 2 = 8 (6 - 2 already placed, not possible)
7th pos = 7 - 2 = 5 (possible)
8th pos = 8 - 2 = 6 (possible)
注:如果我们观察上图，我们会发现 i + k 和 I–k 在第 k 个连续间隔后交替出现。另一个观察结果是，只有当 n 是偶数时，整个排列才能被分成两部分，其中每个部分都必须被 k 整除
C++
// C++ program to find k absolute difference
// permutation
#include<bits/stdc++.h>
using namespace std;
void kDifferencePermutation(int n, int k)
{
// If k is 0 then we just print the
// permutation from 1 to n
if (!k)
{
for (int i = 0; i < n; ++i)
cout << i + 1 << " ";
}
// Check whether permutation is feasible or not
else if (n % (2 * k) != 0)
cout <<"Not Possible";
else
{
for (int i = 0; i < n; ++i)
{
// Put i + k + 1 candidate if position is
// feasible, otherwise put the i - k - 1
// candidate
if ((i / k) % 2 == 0)
cout << i + k + 1 << " ";
else
cout << i - k + 1 << " ";
}
}
cout << "\n";
}
// Driver code
int main()
{
int n = 6 , k = 3;
kDifferencePermutation(n, k);
n = 6 , k = 2;
kDifferencePermutation(n, k);
n = 8 , k = 2;
kDifferencePermutation(n, k);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find k absolute
// difference permutation
import java.io.*;
class GFG {
static void kDifferencePermutation(int n,
int k)
{
// If k is 0 then we just print the
// permutation from 1 to n
if (!(k > 0))
{
for (int i = 0; i < n; ++i)
System.out.print( i + 1 + " ");
}
// Check whether permutation is
// feasible or not
else if (n % (2 * k) != 0)
System.out.print("Not Possible");
else
{
for (int i = 0; i < n; ++i)
{
// Put i + k + 1 candidate
// if position is feasible,
// otherwise put the
// i - k - 1 candidate
if ((i / k) % 2 == 0)
System.out.print( i + k
+ 1 + " ");
else
System.out.print( i - k
+ 1 + " ");
}
}
System.out.println() ;
}
// Driver code
static public void main (String[] args)
{
int n = 6 , k = 3;
kDifferencePermutation(n, k);
n = 6 ;
k = 2;
kDifferencePermutation(n, k);
n = 8 ;
k = 2;
kDifferencePermutation(n, k);
}
}
// This code is contributed by anuj_67.
Python 3
# Python 3 program to find k
# absolute difference permutation
def kDifferencePermutation(n, k):
# If k is 0 then we just print the
# permutation from 1 to n
if (k == 0):
for i in range(n):
print(i + 1, end = " ")
# Check whether permutation
# is feasible or not
elif (n % (2 * k) != 0):
print("Not Possible", end = "")
else:
for i in range(n):
# Put i + k + 1 candidate if position is
# feasible, otherwise put the i - k - 1
# candidate
if (int(i / k) % 2 == 0):
print(i + k + 1, end = " ")
else:
print(i - k + 1, end = " ")
print("\n", end = "")
# Driver code
if __name__ == '__main__':
n = 6
k = 3
kDifferencePermutation(n, k)
n = 6
k = 2
kDifferencePermutation(n, k)
n = 8
k = 2
kDifferencePermutation(n, k)
# This code is contributed by
# Surendra_Gangwar
C
// C# program to find k absolute
// difference permutation
using System;
class GFG {
static void kDifferencePermutation(int n,
int k)
{
// If k is 0 then we just print the
// permutation from 1 to n
if (!(k > 0))
{
for (int i = 0; i < n; ++i)
Console.Write( i + 1 + " ");
}
// Check whether permutation is
// feasible or not
else if (n % (2 * k) != 0)
Console.Write("Not Possible");
else
{
for (int i = 0; i < n; ++i)
{
// Put i + k + 1 candidate
// if position is feasible,
// otherwise put the
// i - k - 1 candidate
if ((i / k) % 2 == 0)
Console.Write( i + k
+ 1 + " ");
else
Console.Write( i - k
+ 1 + " ");
}
}
Console.WriteLine() ;
}
// Driver code
static public void Main ()
{
int n = 6 , k = 3;
kDifferencePermutation(n, k);
n = 6 ;
k = 2;
kDifferencePermutation(n, k);
n = 8 ;
k = 2;
kDifferencePermutation(n, k);
}
}
// This code is contributed by anuj_67.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find k absolute
// difference permutation
function kDifferencePermutation( $n, $k)
{
// If k is 0 then we just print the
// permutation from 1 to n
if (!$k)
{
for($i = 0; $i < $n; ++$i)
echo $i + 1 ," ";
}
// Check whether permutation
// is feasible or not
else if ($n % (2 * $k) != 0)
echo"Not Possible";
else
{
for($i = 0; $i < $n; ++$i)
{
// Put i + k + 1 candidate
// if position is feasible,
// otherwise put the i - k - 1
// candidate
if (($i / $k) % 2 == 0)
echo $i + $k + 1 , " ";
else
echo $i - $k + 1 , " ";
}
}
echo "\n";
}
// Driver Code
$n = 6 ; $k = 3;
kDifferencePermutation($n, $k);
$n = 6 ; $k = 2;
kDifferencePermutation($n, $k);
$n = 8 ;$k = 2;
kDifferencePermutation($n, $k);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// Javascript program to find k absolute difference permutation
function kDifferencePermutation(n, k)
{
// If k is 0 then we just print the
// permutation from 1 to n
if (!(k > 0))
{
for (let i = 0; i < n; ++i)
document.write( i + 1 + " ");
}
// Check whether permutation is
// feasible or not
else if (n % (2 * k) != 0)
document.write("Not Possible");
else
{
for (let i = 0; i < n; ++i)
{
// Put i + k + 1 candidate
// if position is feasible,
// otherwise put the
// i - k - 1 candidate
if (parseInt(i / k, 10) % 2 == 0)
document.write( i + k
+ 1 + " ");
else
document.write( i - k
+ 1 + " ");
}
}
document.write("</br>") ;
}
let n = 6 , k = 3;
kDifferencePermutation(n, k);
n = 6 ;
k = 2;
kDifferencePermutation(n, k);
n = 8 ;
k = 2;
kDifferencePermutation(n, k);
// This code is contributed by rameshtravel07.
</script>
输出:
4 5 6 1 2 3
Not Possible
3 4 1 2 7 8 5 6
时间复杂度:O(n) T3】辅助空间: O(1) 本文由 Shubham Bansal 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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