生成1到N的排列，以使连续数的绝对差给出K个不同的整数

给定两个整数N和K ，其中K ，则任务是生成从1到N的整数排列，以使所有连续整数的绝对差值恰好给出K个不同的整数。

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，递增序列1、2、3，…在偶数索引处，递减序列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 29

C#

// 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 PrinciRaj1992

PHP

= $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
?>

Javascript

 
// Javascript 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
function printPermutation(N, K)
{
     
    // To store the permutation
    var res = [];
    var l = 1, r = N, flag = 0;
 
    for(var i = 0; i < K; i++)
    {
        if (!flag)
        {
             
            // For sequence 1 2 3...
            res.push(l);
            l++;
        }
        else
        {
             
            // For sequence N, N-1, N-2...
            res.push(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(var i = r; i >= l; i--)
            res.push(i);
    }
 
    // If last element added was l - 1
    else
    {
        for(var i = l; i <= r; i++)
            res.push(i);
    }
 
    // Print the permutation
    for(var i = 0; i< res.length; i++)
    {
        document.write(res[i] + " ");
    }
}
 
// Driver code
var N = 10, K = 4;
 
printPermutation(N, K);
 
// This code is contributed by noob2000
 

输出：

1 10 2 9 8 7 6 5 4 3

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