按字母顺序打印给定字符串的所有回文排列
给定一根大小为 n 的绳子绳子。问题是按字母顺序打印串的所有回文排列，如果可能的话打印“-1”。 举例:
Input : str = "aabb"
Output :
abba
baab
Input : malayalam
Output :
aalmymlaa
aamlylmaa
alamymala
almayamla
amalylama
amlayalma
laamymaal
lamayamal
lmaayaaml
maalylaam
malayalam
mlaayaalm
先决条件: 词典第一个回文串和下一个更高的回文号使用同一组数字。 进场:以下为步骤:
- 如果可能，使用字符串的字符找到词典第一个回文字符串，否则打印“-1”并返回。
- 如果从词典学的角度来看，第一个回文字符串被获得，那么打印它。
- 现在，使用字符串的相同字符集找到下一个更高的回文字符串。参考这篇帖子。 这两个帖子唯一的区别是，一个数字在排列，另一个小写字母在排列。
- 如果获得下一个更高的回文字符串，则打印它并转到步骤 3，否则返回。
请注意，字符串字符串总是被操纵，以便使用上述步骤获得回文字符串。
C++
// C++ implementation to print all the palindromic
// permutations alphabetically
#include <bits/stdc++.h>
using namespace std;
const char MAX_CHAR = 26;
// Function to count frequency of each char in the
// string. freq[0] for 'a', ...., freq[25] for 'z'
void countFreq(char str[], int freq[], int n)
{
for (int i = 0; i < n; i++)
freq[str[i] - 'a']++;
}
// Cases to check whether a palindromic
// string can be formed or not
bool canMakePalindrome(int freq[], int n)
{
// count_odd to count no of
// chars with odd frequency
int count_odd = 0;
for (int i = 0; i < MAX_CHAR; i++)
if (freq[i] % 2 != 0)
count_odd++;
// For even length string
// no odd freq character
if (n % 2 == 0) {
if (count_odd > 0)
return false;
else
return true;
}
// For odd length string
// one odd freq character
if (count_odd != 1)
return false;
return true;
}
// Function to find odd freq char and reducing its
// freq by 1, returns garbage value if odd freq
// char is not present
char findOddAndRemoveItsFreq(int freq[])
{
char odd_char;
for (int i = 0; i < MAX_CHAR; i++) {
if (freq[i] % 2 != 0) {
freq[i]--;
odd_char = (char)(i + 'a');
break;
}
}
return odd_char;
}
// To find lexicographically first palindromic
// string.
bool findPalindromicString(char str[], int n)
{
int freq[MAX_CHAR] = { 0 };
countFreq(str, freq, n);
// check whether a palindromic string
// can be formed or not with the
// characters of 'str'
if (!canMakePalindrome(freq, n))
// cannot be formed
return false;
// Assigning odd freq character if present
// else some garbage value
char odd_char = findOddAndRemoveItsFreq(freq);
// indexes to store characters from the front
// and end
int front_index = 0, rear_index = n - 1;
// Traverse characters in alphabetical order
for (int i = 0; i < MAX_CHAR; i++) {
if (freq[i] != 0) {
char ch = (char)(i + 'a');
// store 'ch' to half its frequency times
// from the front and rear end. Note that
// odd character is removed by
// findOddAndRemoveItsFreq()
for (int j = 1; j <= freq[i] / 2; j++) {
str[front_index++] = ch;
str[rear_index--] = ch;
}
}
}
// if true then odd frequency char exists
// store it to its corresponding index
if (front_index == rear_index)
str[front_index] = odd_char;
// palindromic string can be formed
return true;
}
// function to reverse the characters in the
// range i to j in 'str'
void reverse(char str[], int i, int j)
{
while (i < j) {
swap(str[i], str[j]);
i++;
j--;
}
}
// function to find next higher palindromic
// string using the same set of characters
bool nextPalin(char str[], int n)
{
// if length of 'str' is less than '3'
// then no higher palindromic string
// can be formed
if (n <= 3)
return false;
// find the index of last character
// in the 1st half of 'str'
int mid = n / 2 - 1;
int i, j;
// Start from the (mid-1)th character and
// find the first character that is
// alphabetically smaller than the
// character next to it.
for (i = mid - 1; i >= 0; i--)
if (str[i] < str[i + 1])
break;
// If no such character is found, then all characters
// are in descending order (alphabetically) which
// means there cannot be a higher palindromic string
// with same set of characters
if (i < 0)
return false;
// Find the alphabetically smallest character on
// right side of ith character which is greater
// than str[i] up to index 'mid'
int smallest = i + 1;
for (j = i + 2; j <= mid; j++)
if (str[j] > str[i] && str[j] < str[smallest])
smallest = j;
// swap str[i] with str[smallest]
swap(str[i], str[smallest]);
// as the string is a palindrome, the same
// swap of characters should be performed
// in the 2nd half of 'str'
swap(str[n - i - 1], str[n - smallest - 1]);
// reverse characters in the range (i+1) to mid
reverse(str, i + 1, mid);
// if n is even, then reverse characters in the
// range mid+1 to n-i-2
if (n % 2 == 0)
reverse(str, mid + 1, n - i - 2);
// else if n is odd, then reverse characters
// in the range mid+2 to n-i-2
else
reverse(str, mid + 2, n - i - 2);
// next alphabetically higher palindromic
// string can be formed
return true;
}
// function to print all the palindromic permutations
// alphabetically
void printAllPalinPermutations(char str[], int n)
{
// check if lexicographically first palindromic string
// can be formed or not using the characters of 'str'
if (!(findPalindromicString(str, n))) {
// cannot be formed
cout << "-1";
return;
}
// print all palindromic permutations
do {
cout << str << endl;
} while (nextPalin(str, n));
}
// Driver program to test above
int main()
{
char str[] = "malayalam";
int n = strlen(str);
printAllPalinPermutations(str, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to print all the
// palindromic permutations alphabetically
import java.util.*;
class GFG {
static int MAX_CHAR = 26;
// Function to count frequency
// of each char in the string.
// freq[0] for 'a', ...., freq[25] for 'z'
static void countFreq(char str[],
int freq[], int n){
for (int i = 0; i < n; i++)
freq[str[i] - 'a']++;
}
// Cases to check whether a palindromic
// string can be formed or not
static Boolean canMakePalindrome
(int freq[], int n){
// count_odd to count no of
// chars with odd frequency
int count_odd = 0;
for (int i = 0; i < MAX_CHAR; i++)
if (freq[i] % 2 != 0)
count_odd++;
// For even length string
// no odd freq character
if (n % 2 == 0) {
if (count_odd > 0)
return false;
else
return true;
}
// For odd length string
// one odd freq character
if (count_odd != 1)
return false;
return true;
}
// Function for odd freq char and
// reducing its freq by 1, returns
// garbage value if odd freq
// char is not present
static char findOddAndRemoveItsFreq
(int freq[])
{
char odd_char = 'a';
for (int i = 0; i < MAX_CHAR; i++)
{
if (freq[i] % 2 != 0) {
freq[i]--;
odd_char = (char)(i + 'a');
break;
}
}
return odd_char;
}
// To find lexicographically
// first palindromic string.
static boolean findPalindromicString
(char[] str, int n)
{
int[] freq = new int[MAX_CHAR] ;
countFreq(str, freq, n);
// check whether a palindromic
// string can be formed or not
// with the characters of 'str'
if (!canMakePalindrome(freq, n))
return false;
// Assigning odd freq character if
// present else some garbage value
char odd_char =
findOddAndRemoveItsFreq(freq);
// indexes to store characters
// from the front and end
int front_index = 0,
rear_index = n - 1;
// Traverse characters
// in alphabetical order
for (int i = 0; i < MAX_CHAR; i++)
{
if (freq[i] != 0) {
char ch = (char)(i + 'a');
// store 'ch' to half its frequency
// times from the front and rear
// end. Note that odd character is
// removed by findOddAndRemoveItsFreq()
for (int j = 1; j <= freq[i] / 2; j++){
str[front_index++] = ch;
str[rear_index--] = ch;
}
}
}
// if true then odd frequency char exists
// store it to its corresponding index
if (front_index == rear_index)
str[front_index] = odd_char;
// palindromic string can be formed
return true;
}
// function to reverse the characters
// in the range i to j in 'str'
static void reverse(char[] str, int i, int j)
{
char temp;
while (i < j) {
temp = str[i];
str[i] = str[j];
str[j] = temp;
i++;
j--;
}
}
// function to find next higher
// palindromic string using the
// same set of characters
static Boolean nextPalin(char[] str, int n)
{
// if length of 'str' is less
// than '3' then no higher
// palindromic string can be formed
if (n <= 3)
return false;
// find the index of last character
// in the 1st half of 'str'
int mid = n / 2 - 1;
int i, j;
// Start from the (mid-1)th character
// and find the first character
// that is alphabetically smaller
// than the character next to it.
for (i = mid - 1; i >= 0; i--)
if (str[i] < str[i + 1])
break;
// If no such character is found,
// then all characters are in
// descending order (alphabetically)
// which means there cannot be a
// higher palindromic string
// with same set of characters
if (i < 0)
return false;
// Find the alphabetically smallest
// character on right side of ith
// character which is greater than
// str[i] up to index 'mid'
int smallest = i + 1;
for (j = i + 2; j <= mid; j++)
if (str[j] > str[i] && str[j]
< str[smallest])
smallest = j;
char temp;
// swap str[i] with str[smallest]
temp = str[i];
str[i] = str[smallest];
str[smallest] = temp;
// as the string is a palindrome,
// the same swap of characters
// should be performed in the
// 2nd half of 'str'
temp = str[n - i - 1];
str[n - i - 1] = str[n - smallest - 1];
str[n - smallest - 1] = temp;
// reverse characters in the
// range (i+1) to mid
reverse(str, i + 1, mid);
// if n is even, then reverse
// characters in the range
// mid+1 to n-i-2
if (n % 2 == 0)
reverse(str, mid + 1, n - i - 2);
// else if n is odd, then
// reverse characters in
// the range mid+2 to n-i-2
else
reverse(str, mid + 2, n - i - 2);
// next alphabetically higher
// palindromic string can be formed
return true;
}
// function to print all palindromic
// permutations alphabetically
static void printAllPalinPermutations
(char[] str, int n) {
// check if lexicographically
// first palindromic string can
// be formed or not using the
// characters of 'str'
if (!(findPalindromicString(str, n)))
{
// cannot be formed
System.out.print("-1");
return;
}
// print all palindromic permutations
do {
System.out.println(str);
} while (nextPalin(str, n));
}
// Driver program
public static void main(String[] args)
{
char[] str = ("malayalam").toCharArray();
int n = str.length;
printAllPalinPermutations(str, n);
}
}
// This code is contributed by Gitanjali.
Python 3
# Python3 implementation to print all the
# palindromic permutations alphabetically
# import library
import numpy as np
MAX_CHAR = 26
# Function to count frequency of each in the
# string. freq[0] for 'a', ...., freq[25] for 'z'
def countFreq(str, freq, n) :
for i in range(0, n) :
freq[ord(str[i]) - ord('a')] += 1
# Cases to check whether a palindromic
# string can be formed or not
def canMakePalindrome(freq, n) :
# count_odd to count no of
# s with odd frequency
count_odd = 0
for i in range(0, MAX_CHAR) :
if (freq[i] % 2 != 0):
count_odd += 1
# For even length string
# no odd freq character
if (n % 2 == 0):
if (count_odd > 0):
return False
else:
return True
# For odd length string
# one odd freq character
if (count_odd != 1):
return False
return True
# Function to find odd freq and reducing
# its freq by 1, returns garbage
# value if odd freq is not present
def findOddAndRemoveItsFreq(freq) :
odd_char = 0
for i in range(0, MAX_CHAR) :
if (freq[i] % 2 != 0):
freq[i] = freq[i] - 1
odd_char = (chr)(i + ord('a'))
break
return odd_char
# To find lexicographically first
# palindromic string.
def findPalindromicString(str, n) :
freq = np.zeros(26, dtype = np.int)
countFreq(str, freq, n)
# Check whether a palindromic string
# can be formed or not with the
# characters of 'str'
if (not(canMakePalindrome(freq, n))):
# cannot be formed
return False
# Assigning odd freq character if
# present else some garbage value
odd_char = findOddAndRemoveItsFreq(freq)
# indexes to store acters from
# the front and end
front_index = 0; rear_index = n - 1
# Traverse acters in alphabetical order
for i in range(0, MAX_CHAR) :
if (freq[i] != 0) :
ch = (chr)(i + ord('a'))
# Store 'ch' to half its frequency times
# from the front and rear end. Note that
# odd character is removed by
# findOddAndRemoveItsFreq()
for j in range(1, int(freq[i]/2) + 1):
str[front_index] = ch
front_index += 1
str[rear_index] = ch
rear_index -= 1
# if True then odd frequency exists
# store it to its corresponding index
if (front_index == rear_index):
str[front_index] = odd_char
# palindromic string can be formed
return True
# Function to reverse the acters in the
# range i to j in 'str'
def reverse( str, i, j):
while (i < j):
str[i], str[j] = str[j], str[i]
i += 1
j -= 1
# Function to find next higher palindromic
# string using the same set of acters
def nextPalin(str, n) :
# If length of 'str' is less than '3'
# then no higher palindromic string
# can be formed
if (n <= 3):
return False
# Find the index of last acter
# in the 1st half of 'str'
mid = int (n / 2) - 1
# Start from the (mid-1)th acter and
# find the first acter that is
# alphabetically smaller than the
# acter next to it
i = mid -1
while(i >= 0) :
if (str[i] < str[i + 1]):
break
i -= 1
# If no such character is found, then
# all characters are in descending order
# (alphabetically) which means there cannot
# be a higher palindromic string with same
# set of characters
if (i < 0):
return False
# Find the alphabetically smallest character
# on right side of ith character which is
# greater than str[i] up to index 'mid'
smallest = i + 1;
for j in range(i + 2, mid + 1):
if (str[j] > str[i] and str[j] < str[smallest]):
smallest = j
# Swap str[i] with str[smallest]
str[i], str[smallest] = str[smallest], str[i]
# As the string is a palindrome, the same
# swap of characters should be performed
# in the 2nd half of 'str'
str[n - i - 1], str[n - smallest - 1] = str[n - smallest - 1], str[n - i - 1]
# Reverse characters in the range (i+1) to mid
reverse(str, i + 1, mid)
# If n is even, then reverse characters in the
# range mid+1 to n-i-2
if (n % 2 == 0):
reverse(str, mid + 1, n - i - 2)
# else if n is odd, then reverse acters
# in the range mid+2 to n-i-2
else:
reverse(str, mid + 2, n - i - 2)
# Next alphabetically higher palindromic
# string can be formed
return True
# Function to print all the palindromic
# permutations alphabetically
def printAllPalinPermutations(str, n) :
# Check if lexicographically first
# palindromic string can be formed
# or not using the acters of 'str'
if (not(findPalindromicString(str, n))):
# cannot be formed
print ("-1")
return
# Converting list into string
print ( "".join(str))
# print all palindromic permutations
while (nextPalin(str, n)):
# converting list into string
print ("".join(str))
# Driver Code
str= "malayalam"
n = len(str)
# Convertnig string into list so that
# replacement of characters would be easy
str = list(str)
printAllPalinPermutations(str, n)
# This article is contributed by 'saloni1297'
C
// C# code to print all the palindromic
// permutations alphabetically
using System;
class GFG {
static int MAX_CHAR = 26;
// Function to count frequency
// of each char in the string.
// freq[0] for 'a', ...., freq[25] for 'z'
static void countFreq(char []str, int []freq, int n)
{
for (int i = 0; i < n; i++)
freq[str[i] - 'a']++;
}
// Cases to check whether a palindromic
// string can be formed or not
static Boolean canMakePalindrome(int []freq, int n)
{
// count_odd to count no of
// chars with odd frequency
int count_odd = 0;
for (int i = 0; i < MAX_CHAR; i++)
if (freq[i] % 2 != 0)
count_odd++;
// For even length string
// no odd freq character
if (n % 2 == 0) {
if (count_odd > 0)
return false;
else
return true;
}
// For odd length string
// one odd freq character
if (count_odd != 1)
return false;
return true;
}
// Function for odd freq char and
// reducing its freq by 1, returns
// garbage value if odd freq
// char is not present
static char findOddAndRemoveItsFreq(int []freq)
{
char odd_char = 'a';
for (int i = 0; i < MAX_CHAR; i++)
{
if (freq[i] % 2 != 0) {
freq[i]--;
odd_char = (char)(i + 'a');
break;
}
}
return odd_char;
}
// To find lexicographically
// first palindromic string.
static Boolean findPalindromicString(char[] str, int n)
{
int[] freq = new int[MAX_CHAR] ;
countFreq(str, freq, n);
// check whether a palindromic
// string can be formed or not
// with the characters of 'str'
if (!canMakePalindrome(freq, n))
return false;
// Assigning odd freq character if
// present else some garbage value
char odd_char =
findOddAndRemoveItsFreq(freq);
// indexes to store characters
// from the front and end
int front_index = 0,
rear_index = n - 1;
// Traverse characters
// in alphabetical order
for (int i = 0; i < MAX_CHAR; i++)
{
if (freq[i] != 0) {
char ch = (char)(i + 'a');
// store 'ch' to half its frequency
// times from the front and rear
// end. Note that odd character is
// removed by findOddAndRemoveItsFreq()
for (int j = 1; j <= freq[i] / 2; j++){
str[front_index++] = ch;
str[rear_index--] = ch;
}
}
}
// if true then odd frequency char exists
// store it to its corresponding index
if (front_index == rear_index)
str[front_index] = odd_char;
// palindromic string can be formed
return true;
}
// function to reverse the characters
// in the range i to j in 'str'
static void reverse(char[] str, int i, int j)
{
char temp;
while (i < j) {
temp = str[i];
str[i] = str[j];
str[j] = temp;
i++;
j--;
}
}
// function to find next higher
// palindromic string using the
// same set of characters
static Boolean nextPalin(char[] str, int n)
{
// if length of 'str' is less
// than '3' then no higher
// palindromic string can be formed
if (n <= 3)
return false;
// find the index of last character
// in the 1st half of 'str'
int mid = n / 2 - 1;
int i, j;
// Start from the (mid-1)th character
// and find the first character
// that is alphabetically smaller
// than the character next to it.
for (i = mid - 1; i >= 0; i--)
if (str[i] < str[i + 1])
break;
// If no such character is found,
// then all characters are in
// descending order (alphabetically)
// which means there cannot be a
// higher palindromic string
// with same set of characters
if (i < 0)
return false;
// Find the alphabetically smallest
// character on right side of ith
// character which is greater than
// str[i] up to index 'mid'
int smallest = i + 1;
for (j = i + 2; j <= mid; j++)
if (str[j] > str[i] && str[j]
< str[smallest])
smallest = j;
char temp;
// swap str[i] with str[smallest]
temp = str[i];
str[i] = str[smallest];
str[smallest] = temp;
// as the string is a palindrome,
// the same swap of characters
// should be performed in the
// 2nd half of 'str'
temp = str[n - i - 1];
str[n - i - 1] = str[n - smallest - 1];
str[n - smallest - 1] = temp;
// reverse characters in the
// range (i+1) to mid
reverse(str, i + 1, mid);
// if n is even, then reverse
// characters in the range
// mid+1 to n-i-2
if (n % 2 == 0)
reverse(str, mid + 1, n - i - 2);
// else if n is odd, then
// reverse characters in
// the range mid+2 to n-i-2
else
reverse(str, mid + 2, n - i - 2);
// next alphabetically higher
// palindromic string can be formed
return true;
}
// function to print all palindromic
// permutations alphabetically
static void printAllPalinPermutations(char[] str, int n)
{
// check if lexicographically
// first palindromic string can
// be formed or not using the
// characters of 'str'
if (!(findPalindromicString(str, n)))
{
// cannot be formed
Console.WriteLine("-1");
return;
}
// print all palindromic permutations
do {
Console.WriteLine(str);
} while (nextPalin(str, n));
}
// Driver program
public static void Main(String[] args)
{
char[] str = ("malayalam").ToCharArray();
int n = str.Length;
printAllPalinPermutations(str, n);
}
}
// This code is contributed by parashar.
输出:
aalmymlaa
aamlylmaa
alamymala
almayamla
amalylama
amlayalma
laamymaal
lamayamal
lmaayaaml
maalylaam
malayalam
mlaayaalm
时间复杂度:O(mn)，其中 m 是串*的回文排列数。
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