检查重新排列数组元素是否可以形成回文

给定一个大小为N的正整数数组arr ，任务是检查从数组元素的任何排列中形成的数字是否形成回文。

例子：

Input: arr = [1, 2, 3, 1, 2]
Output: Yes
Explanation: The elements of a given array can be rearranged as 1, 2, 3, 2, 1. 
Since 12321 is a palindrome, so output will be “Yes”

Input: arr = [1, 2, 3, 4, 1]
Output: No
Explanation: The elements of a given array cannot be rearranged to form a palindrome within all the possible permutations. So the output will be “No”

方法：给定问题可以使用map来存储数组元素的频率来解决

• 存储所有数组元素的频率
• 检查每个元素的频率是否是偶数
• 对于频率为奇数的元素，如果只有一个这样的元素，则打印 Yes。否则打印编号

下面是上述方法的实现：

C++14

// C++ implementation of the above approach
 
#include 
using namespace std;
 
#define MAX 256
 
// Function to check whether elements of
// an array can form a palindrome
bool can_form_palindrome(int arr[], int n)
{
    // create an empty string
    // to append elements of an array
    string str = "";
 
    // append each element to the string str to form
    // a string so that we can solve it in easy way
    for (int i = 0; i < n; i++) {
        str += arr[i];
    }
 
    // Create a freq array and initialize all
    // values as 0
    int freq[MAX] = { 0 };
 
    // For each character in formed string,
    // increment freq in the corresponding
    // freq array
    for (int i = 0; str[i]; i++) {
        freq[str[i]]++;
    }
    int count = 0;
 
    // Count odd occurring characters
    for (int i = 0; i < MAX; i++) {
        if (freq[i] & 1) {
            count++;
        }
        if (count > 1) {
            return false;
        }
    }
 
    // Return true if odd count is 0 or 1,
    return true;
}
// Drive code
int main()
{
    int arr[] = { 1, 2, 3, 1, 2 };
    int n = sizeof(arr) / sizeof(int);
    can_form_palindrome(arr, n)
        ? cout << "YES"
        : cout << "NO";
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
import java.util.Arrays;
 
class GFG{
 
  static int MAX = 256;
 
  // Function to check whether elements of
  // an array can form a palindrome
  static boolean can_form_palindrome(int []arr, int n)
  {
 
    // create an empty string
    // to append elements of an array
    String str = "";
 
    // append each element to the string str to form
    // a string so that we can solve it in easy way
    for (int i = 0; i < n; i++) {
      str += arr[i];
    }
 
    // Create a freq array and initialize all
    // values as 0
    int freq[] = new int[MAX];
    Arrays.fill(freq,0);
 
    // For each character in formed string,
    // increment freq in the corresponding
    // freq array
    for (int i = 0; i 1) {
        return false;
      }
    }
 
    // Return true if odd count is 0 or 1,
    return true;
  }
 
  // Drive code
  public static void main (String[] args)
  {
    int []arr = { 1, 2, 3, 1, 2 };
    int n = arr.length;
    if(can_form_palindrome(arr, n))
      System.out.println("YES");
    else
      System.out.println("NO");
  }
}
 
// This code is contributed by shivanisinghss2110

Python3

# python implementation of the above approach
 
# Function to check whether elements of
# an array can form a palindrome
def can_form_palindrome(arr, n):
    MAX = 256
    # create an empty string
    # to append elements of an array
    s = ""
 
    # append each element to the string str to form
    # a string so that we can solve it in easy way
    for i in range(n) :
        s = s + str(arr[i])
 
    # Create a freq array and initialize all
    # values as 0
    freq = [0]*MAX
 
    # For each character in formed string,
    # increment freq in the corresponding
    # freq array
    for i in range(N) :
        freq[arr[i]]=freq[arr[i]]+1
     
    count = 0
 
    # Count odd occurring characters
    for i in range(MAX) :
        if (freq[i] & 1) :
            count=count+1
        if (count > 1) :
            return False
         
    # Return true if odd count is 0 or 1,
    return True
   
# Driver Code
if __name__ ==  "__main__":
    arr = [ 1, 2, 3, 1, 2 ]
    N = len(arr)
     
    # Function Call
    if(can_form_palindrome(arr, N)):
        print("YES")
    else:
        print("NO")
                    
 # This code is contributed by anudeep23042002

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
static int MAX = 256;
 
// Function to check whether elements of
// an array can form a palindrome
static bool can_form_palindrome(int []arr, int n)
{
    // create an empty string
    // to append elements of an array
    string str = "";
 
    // append each element to the string str to form
    // a string so that we can solve it in easy way
    for (int i = 0; i < n; i++) {
        str += arr[i];
    }
 
    // Create a freq array and initialize all
    // values as 0
    int []freq = new int[MAX];
    Array.Clear(freq,0,MAX);
 
    // For each character in formed string,
    // increment freq in the corresponding
    // freq array
    for (int i = 0; i 1) {
            return false;
        }
    }
 
    // Return true if odd count is 0 or 1,
    return true;
}
   
// Drive code
public static void Main()
{
    int []arr = { 1, 2, 3, 1, 2 };
    int n = arr.Length;
    if(can_form_palindrome(arr, n))
       Console.Write("YES");
    else
        Console.Write("NO");
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

        // JavaScript Program to implement
        // the above approach
        let MAX = 256
 
        // Function to check whether elements of
        // an array can form a palindrome
        function can_form_palindrome(arr, n)
        {
         
            // create an empty string
            // to append elements of an array
            let str = "";
 
            // append each element to the string str to form
            // a string so that we can solve it in easy way
            for (let i = 0; i < n; i++) {
                str += toString(arr[i]);
            }
 
            // Create a freq array and initialize all
            // values as 0
            let freq = new Array(n).fill(0);
 
            // For each character in formed string,
            // increment freq in the corresponding
            // freq array
            for (let i = 0; i < str.length; i++) {
                freq[str.charCodeAt(i)]++;
            }
            let count = 0;
 
            // Count odd occurring characters
            for (let i = 0; i < MAX; i++) {
                if (freq[i] & 1) {
                    count++;
                }
                if (count > 1) {
                    return false;
                }
            }
 
            // Return true if odd count is 0 or 1,
            return true;
        }
         
        // Drive code
        let arr = [1, 2, 3, 1, 2];
        let n = arr.length;
        can_form_palindrome(arr, n)
            ? document.write("YES")
            : document.write("NO");
 
// This code is contributed by Potta Lokesh
    

输出

YES

时间复杂度： O(N)
辅助空间： O(N)