给定一个字符串str，找到其所有回文排列的计数。
例子 ：
Input : str = "gfgf"
Output : 2
There are two palindromic
permutations fggf and gffg
Input : str = "abc"
Output : 0
这个想法是基于以下事实：
- 如果出现的奇数个字符最多为1个，则字符串可以置换为回文。
- 唯一出现的唯一字符总是出现在中间。
- 所有字符计数的一半决定结果。在出现奇数字符的情况下，它是下限的一半。另一半自动决定
例如，如果输入字符串为“ aabbccd”，则回文排列数为3！ (我们将所有计数的一半作为下限获得三)
如果一半本身有重复的字符怎么办?
我们应用简单的组合规则，并除以一半。
例如，“ aaaaaabbbb”，字符串的一半为5。在回文字符串的一半中，“ a”重复3次，“ b”重复2次，所以我们的结果是(5！)/(2！)。 *(3！)。
C++
// CPP program to find number of
// palindromic permutations of a
// given string
#include
using namespace std;
const int MAX = 256;
// Returns factorial of n
long long int fact(int n)
{
long long int res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
// Returns count of palindromic
// permutations of str.
int countPalinPermutations(string &str)
{
// Count frequencies of all characters
int n = str.length();
int freq[MAX] = { 0 };
for (int i = 0; i < n; i++)
freq[str[i]]++;
// Since half of the characters
// decide count of palindromic
// permutations, we take (n/2)!
long long int res = fact(n / 2);
// To make sure that there is at
// most one odd occurring char
bool oddFreq = false;
// Traverse through all counts
for (int i = 0; i < MAX; i++) {
int half = freq[i] / 2;
// To make sure that the
// string can permute to
// form a palindrome
if (freq[i] % 2 != 0) {
// If there are more than
// one odd occurring chars
if (oddFreq == true)
return 0;
oddFreq = true;
}
// Divide all permutations with
// repeated characters
res = res / fact(half);
}
return res;
}
// Driver code
int main()
{
string str = "gffg";
cout << countPalinPermutations(str);
return 0;
} Java
// Java program to find number of
// palindromic permutations of a
// given string
class GFG {
static final int MAX = 256;
// Returns factorial of n
static long fact(int n)
{
long res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
// Returns count of palindromic
// permutations of str.
static int countPalinPermutations(String str)
{
// Count frequencies of all characters
int n = str.length();
int freq[]=new int[MAX];
for (int i = 0; i < n; i++)
freq[str.charAt(i)]++;
// Since half of the characters
// decide count of palindromic
// permutations, we take (n/2)!
long res = fact(n / 2);
// To make sure that there is at
// most one odd occurring char
boolean oddFreq = false;
// Traverse through all counts
for (int i = 0; i < MAX; i++) {
int half = freq[i] / 2;
// To make sure that the
// string can permute to
// form a palindrome
if (freq[i] % 2 != 0) {
// If there are more than
// one odd occurring chars
if (oddFreq == true)
return 0;
oddFreq = true;
}
// Divide all permutations with
// repeated characters
res = res / fact(half);
}
return (int)res;
}
// Driver code
public static void main (String[] args)
{
String str = "gffg";
System.out.print(
countPalinPermutations(str));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python3 program to find number of
# palindromic permutations of a
# given string
MAX = 256
# Returns factorial of n
def fact(n) :
res = 1
for i in range(2, n+1) :
res = res * i
return res
# Returns count of palindromic
# permutations of str.
def countPalinPermutations(str) :
global MAX
# Count frequencies of
# all characters
n = len(str)
freq = [0] * MAX;
for i in range(0, n) :
freq[ord(str[i])] = freq[ord(str[i])] + 1;
# Since half of the characters
# decide count of palindromic
# permutations, we take (n/2)!
res = fact(int(n / 2))
# To make sure that there is at
# most one odd occurring char
oddFreq = False
# Traverse through all counts
for i in range(0, MAX) :
half = int(freq[i] / 2)
# To make sure that the
# string can permute to
# form a palindrome
if (freq[i] % 2 != 0):
# If there are more than
# one odd occurring chars
if (oddFreq == True):
return 0
oddFreq = True
# Divide all permutations
# with repeated characters
res = int(res / fact(half))
return res
# Driver code
str = "gffg"
print (countPalinPermutations(str))
# This code is contributed by Manish Shaw
# (manishshaw1)C#
// C# program to find number of
// palindromic permutations of a
// given string
using System;
class GFG {
static int MAX = 256;
// Returns factorial of n
static long fact(int n)
{
long res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
// Returns count of palindromic
// permutations of str.
static int countPalinPermutations(string str)
{
// Count frequencies of all characters
int n = str.Length;
int []freq=new int[MAX];
for (int i = 0; i < n; i++)
freq[str[i]]++;
// Since half of the characters
// decide count of palindromic
// permutations, we take (n/2)!
long res = fact(n / 2);
// To make sure that there is at
// most one odd occurring char
bool oddFreq = false;
// Traverse through all counts
for (int i = 0; i < MAX; i++) {
int half = freq[i] / 2;
// To make sure that the
// string can permute to
// form a palindrome
if (freq[i] % 2 != 0) {
// If there are more than
// one odd occurring chars
if (oddFreq == true)
return 0;
oddFreq = true;
}
// Divide all permutations with
// repeated characters
res = res / fact(half);
}
return (int)res;
}
// Driver code
public static void Main ()
{
string str = "gffg";
Console.WriteLine(
countPalinPermutations(str));
}
}
// This code is contributed by vt_m.PHP
输出 ：
2
上面的解决方案很早就导致溢出。我们可以通过执行模块化算术来避免溢出。在下一篇文章中，我们将讨论基于模块化算术的方法。