查找包含总和小于按位异或的子数组的 N 长度排列

给定一个正整数N ，任务是找到一个长度为N的排列，它的任何子数组的按位 OR 大于或等于子数组的长度。

例子：

Input: N = 5 
Output: 1 3 5 2 4 
Explanation: 
Consider the subarray {1, 3, 5} from the permutation {1, 3, 5, 2, $}. 
Length of the subarray = 3 
Bitwise OR of the subarray = 1 | 3 | 5 = 7 ( which is greater than 3) 
Similarly, every subarray of any length smaller than equal to N will satisfy the condition.

Input: 4 
Output: 4 3 1 2

方法：实际上任何长度N 的排列都满足要求的条件，基于以下事实：

• 任何一组数字的按位 OR 大于或等于该组中存在的最大数字。
• 由于任何子数组将包含至少一个大于或等于其长度的元素，因此任何排列都满足给定条件。

下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
#include 
using namespace std;
 
// Function to print the
// required permutation
void findPermutation(int N)
{
   for(int i = 1; i <= N; i++)
     cout << i << " ";
   
   cout << endl;
}
 
// Driver code
int main()
{
 
    int N = 5;
   
    findPermutation(N);
   
    return 0;
}

Java

// Java implementation of
// the above approach
import java.util.*;
 
class GFG{
     
// Function to print the
// required permutation
static void findPermutation(int N)
{
    for(int i = 1; i <= N; i++)
        System.out.print(i + " ");
}
     
// Driver Code
public static void main(String[] args)
{
     int N = 5;
    
    findPermutation(N);
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python3 implementation of
# the above approach
 
# Function to print the
# required permutation
def findPermutation(N):
    for i in range(1, N + 1, 1):
        print(i, end= " ") 
    print("\n", end = "")
 
# Driver code
if __name__ == '__main__':
    N = 5
    findPermutation(N)
     
    # This code is contributed by ipg2016107.

C#

// C# implementation of
// the above approach
using System;
 
class GFG{
  
// Function to print the
// required permutation
static void findPermutation(int N)
{
    for(int i = 1; i <= N; i++)
        Console.Write(i + " ");
}
 
// Driver code
public static void Main()
{
    int N = 5;
   
    findPermutation(N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

// javascript implementation of
// the above approach
 
// Function to print the
// required permutation
function findPermutation(N)
{
   for(var i = 1; i <= N; i++)
     document.write( i + " ");
   
   document.write("<br>");
}
 
var N = 5;
findPermutation(N);
 
//This code in contributed by SoumikMondal

输出：

1 2 3 4 5

时间复杂度： O(N)
辅助空间： O(1)