dsal 前 N 个自然数排列中的第 k 个元素，所有偶数以递增顺序放在奇数之前

前 N 个自然数排列中的第 k 个元素，所有偶数以递增顺序放在奇数之前

原文:https://www . geesforgeks . org/kth-第一个 n 个自然数排列中的元素-将所有偶数放在奇数之前-递增顺序/

给定两个整数 N 和 K ，任务是在排列的第一个NT10】自然数的排列中找到第 K ^第T7】元素，使得所有偶数以递增的顺序出现在奇数之前。

示例:

输入: N = 10，K = 3 输出: 6 说明: 需要的排列是{2，4，6，8，10，1，3，5，7，9}。排列中的 3 ^rd 数为 6。

输入: N = 5，K = 4 输出: 3 说明: 需要的排列是{2，4，1，3，5}。排列中的第 4 个^{数是 3。}

天真方法:解决问题最简单的方法是首先生成所需的排列的 N 自然数，然后遍历排列找到其中存在的 K ^th 元素。按照以下步骤解决问题:

初始化一个数组，说 V[] 大小 N 。，以存储所需的序列。
将所有小于或等于 N 的偶数插入 V[] 。
然后，将所有小于或等于 N 的奇数插入 V[] 。
形成数组后，打印V【K–1】的值作为结果。

下面是上述方法的实现:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the K-th element
// in the required permutation
void findKthElement(int N, int K)
{
    // Stores the required permutation
    vector<int> v;

    // Insert all the even numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
        if (i % 2 == 0) {
            v.push_back(i);
        }
    }

    // Now, insert all odd numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
        if (i % 2 != 0) {
            v.push_back(i);
        }
    }

    // Print the Kth element
    cout << v[K - 1];
}

// Driver Code
int main()
{
    int N = 10, K = 3;
    findKthElement(N, K);

    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{

  // Function to find the K-th element
  // in the required permutation
  static void findKthElement(int N, int K)
  {

    // Stores the required permutation
    ArrayList<Integer> v = new ArrayList<>();

    // Insert all the even numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
      if (i % 2 == 0) {
        v.add(i);
      }
    }

    // Now, insert all odd numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
      if (i % 2 != 0) {
        v.add(i);
      }
    }

    // Print the Kth element
    System.out.println(v.get(K - 1));
  }

  // Driver code
  public static void main(String[] args)
  {

    int N = 10, K = 3;

    // functions call
    findKthElement(N, K);
  }
}

// This code is contributed by Kingash.

Python 3

# python 3 program for the above approach

# Function to find the K-th element
# in the required permutation
def findKthElement(N, K):

    # Stores the required permutation
    v = []

    # Insert all the even numbers
    # less than or equal to N
    for i in range(1, N + 1):
        if (i % 2 == 0):
            v.append(i)

    # Now, insert all odd numbers
    # less than or equal to N
    for i in range(1, N + 1):
        if (i % 2 != 0):
            v.append(i)

    # Print the Kth element
    print(v[K - 1])

# Driver Code
if __name__ == "__main__":
    N = 10
    K = 3
    findKthElement(N, K)

    # This code is contributed by ukasp.

C

// C# program for above approach
using System;
using System.Collections.Generic;

public class GFG
{

  // Function to find the K-th element
  // in the required permutation
  static void findKthElement(int N, int K)
  {

    // Stores the required permutation
    List<int> v = new List<int>();

    // Insert all the even numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
      if (i % 2 == 0) {
        v.Add(i);
      }
    }

    // Now, insert all odd numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
      if (i % 2 != 0) {
        v.Add(i);
      }
    }

    // Print the Kth element
    Console.WriteLine(v[K - 1]);
  }

  // Driver code
  public static void Main(String[] args)
  {
    int N = 10, K = 3;

    // functions call
    findKthElement(N, K);
  }
}

// This code is contributed by susmitakundugoaldanga.

java 描述语言

<script>

// JavaScript program for the above approach

    // Function to find the K-th element
    // in the required permutation
    function findKthElement(N , K) {

        // Stores the required permutation
        var v = [];

        // Insert all the even numbers
        // less than or equal to N
        for (i = 1; i <= N; i++) {
            if (i % 2 == 0) {
                v.push(i);
            }
        }

        // Now, insert all odd numbers
        // less than or equal to N
        for (i = 1; i <= N; i++) {
            if (i % 2 != 0) {
                v.push(i);
            }
        }

        // Print the Kth element
        document.write(v[K - 1]);
    }

    // Driver code

        var N = 10, K = 3;

        // functions call
        findKthElement(N, K);

// This code contributed by Rajput-Ji

</script>

Output

时间复杂度:O(N) T5辅助空间:** O(N)

高效进场:对上述进场进行优化，思路是基于前半段第一 N / 2 元素为偶数，K^thT7】元素的值等于 K * 2 。如果 K > N/2 ， K ^th 元素的值取决于 N 是奇数还是偶数。按照以下步骤解决问题:

初始化一个变量，说 ans，来存储K^thT5】元素。
检查 K 值是否≤ N/2 。如果发现属实，将 ans 更新为 K*2 。
否则 K 位于后半段。在这种情况下， ans 取决于 N 的值。
- 如果 N 的值为偶数，则将 ans 更新为 (K*2)-N-1 。
- 否则，将 ans 更新为 (K*2)-N 。
打印和的值作为结果。

下面是上述方法的实现:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the Kth element
// in the required permutation
void findKthElement(int N, int K)
{
    // Store the required result
    int ans = 0;

    // If K is in the first
    // N / 2 elements, print K * 2
    if (K <= N / 2) {
        ans = K * 2;
    }

    // Otherwise, K is greater than N/2
    else {

        // If N is even
        if (N % 2 == 0) {
            ans = (K * 2) - N - 1;
        }

        // If N is odd
        else {
            ans = (K * 2) - N;
        }
    }

    // Print the required result
    cout << ans;
}

// Driver Code
int main()
{
    int N = 10, K = 3;
    findKthElement(N, K);

    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {

  // Function to find the Kth element
  // in the required permutation
  static void findKthElement(int N, int K)
  {
    // Store the required result
    int ans = 0;

    // If K is in the first
    // N / 2 elements, print K * 2
    if (K <= N / 2) {
      ans = K * 2;
    }

    // Otherwise, K is greater than N/2
    else {

      // If N is even
      if (N % 2 == 0) {
        ans = (K * 2) - N - 1;
      }

      // If N is odd
      else {
        ans = (K * 2) - N;
      }
    }

    // Print the required result
    System.out.println(ans);
  }

  // Driver code
  public static void main(String[] args)
  {

    int N = 10, K = 3;

    // functions call
    findKthElement(N, K);
  }
}

// This code is contributed by Kingash.

Python 3

# Python 3 program for the above approach

# Function to find the Kth element
# in the required permutation
def findKthElement(N, K):

    # Store the required result
    ans = 0

    # If K is in the first
    # N / 2 elements, print K * 2
    if (K <= N / 2):
        ans = K * 2

    # Otherwise, K is greater than N/2
    else:

        # If N is even
        if (N % 2 == 0):
            ans = (K * 2) - N - 1

        # If N is odd
        else:
            ans = (K * 2) - N

    # Print the required result
    print(ans)

# Driver Code
if __name__ == '__main__':
    N = 10
    K = 3
    findKthElement(N, K)

    # This code is contributed by ipg2016107.

C

// C# program for the above approach
using System;

class GFG{

  // Function to find the Kth element
  // in the required permutation
  static void findKthElement(int N, int K)
  {

    // Store the required result
    int ans = 0;

    // If K is in the first
    // N / 2 elements, print K * 2
    if (K <= N / 2) {
      ans = K * 2;
    }

    // Otherwise, K is greater than N/2
    else {

      // If N is even
      if (N % 2 == 0) {
        ans = (K * 2) - N - 1;
      }

      // If N is odd
      else {
        ans = (K * 2) - N;
      }
    }

    // Print the required result
    Console.Write(ans);
  }

  // Driver code
  static void Main()
  {
    int N = 10, K = 3;

    // functions call
    findKthElement(N, K);
  }
}

// This code is contributed by sanjoy_62.

java 描述语言

<script>
// javascript program for the above approach

// Function to find the Kth element
// in the required permutation
function findKthElement( N,  K)
{

    // Store the required result
    let ans = 0;

    // If K is in the first
    // N / 2 elements, print K * 2
    if (K <= N / 2)
    {
        ans = K * 2;
    }

    // Otherwise, K is greater than N/2
    else
    {

        // If N is even
        if (N % 2 == 0)
        {
            ans = (K * 2) - N - 1;
        }

        // If N is odd
        else
        {
            ans = (K * 2) - N;
        }
    }

    // Print the required result
    document.write(ans);
}

// Driver Code
    let N = 10, K = 3;
    findKthElement(N, K);

// This code is contributed by todaysgaurav
</script>

Output

时间复杂度:O(1) T5辅助空间:** O(1)

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前 N 个自然数排列中的第 k 个元素，所有偶数以递增顺序放在奇数之前

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dsal 前 N 个自然数排列中的第 k 个元素，所有偶数以递增顺序放在奇数之前

前 N 个自然数排列中的第 k 个元素，所有偶数以递增顺序放在奇数之前

C++

Java 语言(一种计算机语言，尤用于创建网站)

Python 3

C

java 描述语言

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Java 语言(一种计算机语言，尤用于创建网站)

Python 3

C

java 描述语言

留言

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