打印字符串的所有回文排列

给定一个字符串，我们需要打印可以使用该字符串的字母生成的所有可能的回文。

例子：

Input:  str = "aabcb"
Output: abcba bacab

Input:  str = "aabbcadad"
Output: aabdcdbaa aadbcbdaa abadcdaba
        abdacadba adabcbada adbacabda
        baadcdaab badacadab bdaacaadb
        daabcbaad dabacabad dbaacaabd

回文的生成可以通过以下步骤完成，

1. 首先我们需要检查字符串的字母是否可以构成回文，如果不是则返回。
2. 经过上述检查，我们可以通过取给定字符串的每个字母的一半频率来制作第一个回文字符串的一半（按字典顺序最小）。
3. 现在遍历这半字符串的所有可能排列，每次在末尾添加这部分的反转，如果字符串长度为奇数，则在中间添加奇数频率字符，以制作回文。

下面是 C++ 实现。

C

// C++ program to print all palindrome permutations of
// a string.
#include 
using namespace std;
#define M 26
  
/* Utility function to count frequencies and checking
    whether letter can make a palindrome or not */
bool isPalin(string str, int* freq)
{
    /* Initialising frequency array with all zeros */
    memset(freq, 0, M * sizeof(int));
    int l = str.length();
  
    /* Updating frequency according to given string */
    for (int i = 0; i < l; i++)
        freq[str[i] - 'a']++;
  
    int odd = 0;
  
    /* Loop to count total letter with odd frequency */
    for (int i = 0; i < M; i++)
        if (freq[i] % 2 == 1)
            odd++;
  
    /* Palindrome condition :
    if length is odd then only one letter's frequency must be odd
    if length is even no letter should have odd frequency */
    if ((l % 2 == 1 && odd == 1 ) || (l %2 == 0 && odd == 0))
        return true;
    else
        return false;
}
  
/* Utility function to reverse a string */
string reverse(string str)
{
    string rev = str;
    reverse(rev.begin(), rev.end());
    return rev;
}
  
/* Function to print all possible palindromes by letter of
    given string */
void printAllPossiblePalindromes(string str)
{
    int freq[M];
  
    // checking whether letter can make palindrome or not
    if (!isPalin(str, freq))
        return;
  
    int l = str.length();
  
    // half will contain half part of all palindromes,
    // that is why pushing half freq of each letter
    string half = "";
    char oddC;
    for (int i = 0; i < M; i++)
    {
        /* This condition will be true at most once */
        if(freq[i] % 2 == 1)
            oddC = i + 'a';
  
        half += string(freq[i] / 2, i + 'a');
    }
  
    /* palin will store the possible palindromes one by one */
    string palin;
  
    // Now looping through all permutation of half, and adding
    // reverse part at end.
    // if length is odd, then pushing oddCharacter also in mid
    do
    {
        palin = half;
        if (l % 2 == 1)
            palin += oddC;
        palin += reverse(half);
        cout << palin << endl;
    }
    while (next_permutation(half.begin(), half.end()));
}
  
// Driver Program to test above function
int main()
{
    string str = "aabbcadad";
    cout << "All palindrome permutations of " << str << endl;
    printAllPossiblePalindromes(str);
    return 0;
}

输出：

All palindrome permutations of aabbcadad
aabdcdbaa
aadbcbdaa
abadcdaba
abdacadba
adabcbada
adbacabda
baadcdaab
badacadab
bdaacaadb
daabcbaad
dabacabad
dbaacaabd

插图 ：

Let given string is "aabbcadad"

Letters have following frequencies : 
 a(4), b(2), c(1), d(2).

As all letter has even frequency except one we can 
make palindromes with the letter of this string.

Now half part is – aabd

So traversing through all possible permutations of 
this half string and adding odd frequency character 
and reverse of string at the end we get following 
possible palindrome as final result : 
aabdcdbaa   aadbcbdaa   abadcdaba   abdacadba
adabcbada   adbacabda   baadcdaab   badacadab
bdaacaadb   daabcbaad   dabacabad   dbaacaabd