检查字符串的任何字谜是否为回文

我们给出了一个字谜字符串，我们必须检查它是否可以变成回文。
例子：

Input : geeksforgeeks 
Output : No
There is no palindrome anagram of 
given string

Input  : geeksgeeks
Output : Yes
There are palindrome anagrams of
given string. For example kgeesseegk

这个问题基本上是一样的检查，如果给定的字符串的字符可以重新排列成回文。我们可以使用计数数组在 O(n) 时间内完成。以下是详细步骤。
1) 创建一个字母大小的计数数组，通常为 256。将计数数组的所有值初始化为 0。
2) 遍历给定的字符串并递增每个字符的计数。
3) 遍历count数组，如果count数组有多个奇数，则返回false。否则，返回真。

C++

#include 
using namespace std;
#define NO_OF_CHARS 256
 
/* function to check whether characters of a string
   can form a palindrome */
bool canFormPalindrome(string str)
{
    // Create a count array and initialize all
    // values as 0
    int count[NO_OF_CHARS] = { 0 };
 
    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for (int i = 0; str[i]; i++)
        count[str[i]]++;
 
    // Count odd occurring characters
    int odd = 0;
    for (int i = 0; i < NO_OF_CHARS; i++) {
        if (count[i] & 1)
            odd++;
 
        if (odd > 1)
            return false;
    }
 
    // Return true if odd count is 0 or 1,
    return true;
}
 
/* Driver program to test to print printDups*/
int main()
{
    canFormPalindrome("geeksforgeeks") ? cout << "Yes\n" : cout << "No\n";
    canFormPalindrome("geeksogeeks") ? cout << "Yes\n" : cout << "No\n";
    return 0;
}

Java

// Java program to Check if any anagram
// of a string is palindrome or not
public class GFG {
    static final int NO_OF_CHARS = 256;
 
    /* function to check whether characters of
      a string can form a palindrome */
    static boolean canFormPalindrome(String str)
    {
        // Create a count array and initialize
        // all values as 0
        int[] count = new int[NO_OF_CHARS];
 
        // For each character in input strings,
        // increment count in the corresponding
        // count array
        for (int i = 0; i < str.length(); i++)
            count[str.charAt(i)]++;
 
        // Count odd occurring characters
        int odd = 0;
        for (int i = 0; i < NO_OF_CHARS; i++) {
            if ((count[i] & 1) != 0)
                odd++;
 
            if (odd > 1)
                return false;
        }
 
        // Return true if odd count is 0 or 1,
        return true;
    }
 
    /* Driver program to test to print printDups*/
    public static void main(String args[])
    {
        System.out.println(canFormPalindrome("geeksforgeeks")
                               ? "Yes"
                               : "No");
        System.out.println(canFormPalindrome("geeksogeeks")
                               ? "Yes"
                               : "No");
    }
}
// This code is contributed by Sumit Ghosh

Python

NO_OF_CHARS = 256
   
""" function to check whether characters of a string
   can form a palindrome """
def canFormPalindrome(string):
     
    # Create a count array and initialize all
    # values as 0
    count = [0 for i in range(NO_OF_CHARS)]
   
    # For each character in input strings,
    # increment count in the corresponding
    # count array
    for i in string:
        count[ord(i)] += 1
   
    # Count odd occurring characters
    odd = 0
    for i in range(NO_OF_CHARS):
        if (count[i] & 1):
            odd += 1
  
        if (odd > 1):
            return False
   
    # Return true if odd count is 0 or 1,
    return True
   
# Driver program to test to print printDups
if(canFormPalindrome("geeksforgeeks")):
    print "Yes"
else:
    print "No"
if(canFormPalindrome("geeksogeeks")):
    print "Yes"
else:
    print "NO"
 
# This code is contributed by Sachin Bisht

C#

// C# program to Check if any anagram
// of a string is palindrome or not
using System;
 
public class GFG {
     
    static int NO_OF_CHARS = 256;
 
    /* function to check whether
    characters of a string can form
    a palindrome */
    static bool canFormPalindrome(string str)
    {
         
        // Create a count array and
        // initialize all values as 0
        int[] count = new int[NO_OF_CHARS];
 
        // For each character in input
        // strings, increment count in
        // the corresponding count array
        for (int i = 0; i < str.Length; i++)
            count[str[i]]++;
 
        // Count odd occurring characters
        int odd = 0;
        for (int i = 0; i < NO_OF_CHARS; i++) {
            if ((count[i] & 1) != 0)
                odd++;
 
            if (odd > 1)
                return false;
        }
 
        // Return true if odd count
        // is 0 or 1,
        return true;
    }
 
    // Driver program
    public static void Main()
    {
        Console.WriteLine(
            canFormPalindrome("geeksforgeeks")
                              ? "Yes" : "No");
                               
        Console.WriteLine(
            canFormPalindrome("geeksogeeks")
                              ? "Yes" : "No");
    }
}
 
// This code is contributed by vt_m.

Javascript

    // Javascript program to Check if any anagram
    // of a string is palindrome or not
     
    let NO_OF_CHARS = 256;
  
    /* function to check whether
    characters of a string can form
    a palindrome */
    function canFormPalindrome(str)
    {
          
        // Create a count array and
        // initialize all values as 0
        let count = new Array(NO_OF_CHARS);
        count.fill(0);
  
        // For each character in input
        // strings, increment count in
        // the corresponding count array
        for (let i = 0; i < str.length; i++)
            count[str[i].charCodeAt()]++;
  
        // Count odd occurring characters
        let odd = 0;
        for (let i = 0; i < NO_OF_CHARS; i++) {
            if ((count[i] & 1) != 0)
                odd++;
  
            if (odd > 1)
                return false;
        }
  
        // Return true if odd count
        // is 0 or 1,
        return true;
    }
     
    document.write(canFormPalindrome("geeksforgeeks")? "Yes" + "</br>" : "No" + "</br>");
                                
      document.write(canFormPalindrome("geeksogeeks")? "Yes" : "No");
 
// This code is contributed by divyeshrabadiya07.

输出：

No
Yes

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