给定一个字符串，计算存在多少个最大长度的回文(不必是子字符串)
例子：
Input : str = "ababa"
Output: 2
Explanation :
palindromes of maximum of lenghts are :
"ababa", "baaab"
Input : str = "ababab"
Output: 4
Explanation :
palindromes of maximum of lenghts are :
"ababa", "baaab", "abbba", "babab"
方法回文可以表示为“ str + t + reverse(str)” 。
注意：对于长度相等的回文字符串， “ t”为空
计算可以用多少种方式形成“ str”，然后乘以“ t”(省略的单个字符数)。
令c i为字符串中字符出现的次数。考虑以下情况：
1.如果c i是偶数。那么每个最大回文集的一半将恰好包含字母f i = c i / 2。
2.如果c i是奇数。然后，每个最大回文集的一半将完全包含字母f i =(c i – 1)/ 2。
令k为奇数c i的数量。如果k = 0，则最大回文长度将是偶数；否则它将是奇数，并且将恰好有k个可能的中间字母，即我们可以将该字母设置为回文的中间。
n个对象与n 1个相同类型1的对象，n 2个相同类型2的对象以及n3个类型3的相同对象的排列数目为n！ /(n1！* n2！* n3！) 。
因此，这里的字符总数为f a + f b + f a +……。+ f y + f z 。因此，排列数为(f a + f b + f a +……。+ f y + f z )！ /一！ f b ！…f y ！f z ！。
现在，如果K不为0，答案显然就是k *(f a + f b + f a +……。+ f y + f z +)！ /一！ f b ！…f y ！f z ！
下面是以上的实现。
C++
// C++ implementation for counting
// maximum length palindromes
#include
using namespace std;
// factorial of a number
int fact(int n)
{
int ans = 1;
for (int i = 1; i <= n; i++)
ans = ans * i;
return (ans);
}
// function to count maximum length palindromes.
int numberOfPossiblePallindrome(string str, int n)
{
// Count number of occurrence
// of a charterer in the string
unordered_map mp;
for (int i = 0; i < n; i++)
mp[str[i]]++;
int k = 0; // Count of singles
int num = 0; // numerator of result
int den = 1; // denominator of result
int fi;
for (auto it = mp.begin(); it != mp.end(); ++it)
{
// if frequency is even
// fi = ci / 2
if (it->second % 2 == 0)
fi = it->second / 2;
// if frequency is odd
// fi = ci - 1 / 2.
else
{
fi = (it->second - 1) / 2;
k++;
}
// sum of all frequencies
num = num + fi;
// product of factorial of
// every frequency
den = den * fact(fi);
}
// if all character are unique
// so there will be no pallindrome,
// so if num != 0 then only we are
// finding the factorial
if (num != 0)
num = fact(num);
int ans = num / den;
if (k != 0)
{
// k are the single
// elements that can be
// placed in middle
ans = ans * k;
}
return (ans);
}
// Driver program
int main()
{
char str[] = "ababab";
int n = strlen(str);
cout << numberOfPossiblePallindrome(str, n);
return 0;
} Java
// Java implementation for counting
// maximum length palindromes
import java.util.*;
class GFG
{
// factorial of a number
static int fact(int n)
{
int ans = 1;
for (int i = 1; i <= n; i++)
ans = ans * i;
return (ans);
}
// function to count maximum length palindromes.
static int numberOfPossiblePallindrome(String str, int n)
{
// Count number of occurrence
// of a charterer in the string
Map mp = new HashMap<>();
for (int i = 0; i < n; i++)
mp.put( str.charAt(i),mp.get( str.charAt(i))==null?
1:mp.get( str.charAt(i))+1);
int k = 0; // Count of singles
int num = 0; // numerator of result
int den = 1; // denominator of result
int fi;
for (Map.Entry it : mp.entrySet())
{
// if frequency is even
// fi = ci / 2
if (it.getValue() % 2 == 0)
fi = it.getValue() / 2;
// if frequency is odd
// fi = ci - 1 / 2.
else
{
fi = (it.getValue() - 1) / 2;
k++;
}
// sum of all frequencies
num = num + fi;
// product of factorial of
// every frequency
den = den * fact(fi);
}
// if all character are unique
// so there will be no pallindrome,
// so if num != 0 then only we are
// finding the factorial
if (num != 0)
num = fact(num);
int ans = num / den;
if (k != 0)
{
// k are the single
// elements that can be
// placed in middle
ans = ans * k;
}
return (ans);
}
// Driver code
public static void main(String[] args)
{
String str = "ababab";
int n = str.length();
System.out.println(numberOfPossiblePallindrome(str, n));
}
}
// This code is contributed by Princi Singh Python3
# Python3 implementation for counting
# maximum length palindromes
import math as mt
# factorial of a number
def fact(n):
ans = 1
for i in range(1, n + 1):
ans = ans * i
return (ans)
# function to count maximum length palindromes.
def numberOfPossiblePallindrome(string, n):
# Count number of occurrence
# of a charterer in the string
mp = dict()
for i in range(n):
if string[i] in mp.keys():
mp[string[i]] += 1
else:
mp[string[i]] = 1
k = 0 # Count of singles
num = 0 # numerator of result
den = 1 # denominator of result
fi = 0
for it in mp:
# if frequency is even
# fi = ci / 2
if (mp[it] % 2 == 0):
fi = mp[it] // 2
# if frequency is odd
# fi = ci - 1 / 2.
else:
fi = (mp[it] - 1) // 2
k += 1
# sum of all frequencies
num = num + fi
# product of factorial of
# every frequency
den = den * fact(fi)
# if all character are unique
# so there will be no pallindrome,
# so if num != 0 then only we are
# finding the factorial
if (num != 0):
num = fact(num)
ans = num //den
if (k != 0):
# k are the single
# elements that can be
# placed in middle
ans = ans * k
return (ans)
# Driver Code
string = "ababab"
n = len(string)
print(numberOfPossiblePallindrome(string, n))
# This code is contributed by
# Mohit kumar 29C#
// C# implementation for counting
// maximum length palindromes
using System;
using System.Collections.Generic;
class GFG
{
// factorial of a number
static int fact(int n)
{
int ans = 1;
for (int i = 1; i <= n; i++)
ans = ans * i;
return (ans);
}
// function to count maximum length palindromes.
static int numberOfPossiblePallindrome(String str, int n)
{
// Count number of occurrence
// of a charterer in the string
Dictionary mp = new Dictionary();
for (int i = 0 ; i < n; i++)
{
if(mp.ContainsKey(str[i]))
{
var val = mp[str[i]];
mp.Remove(str[i]);
mp.Add(str[i], val + 1);
}
else
{
mp.Add(str[i], 1);
}
}
int k = 0; // Count of singles
int num = 0; // numerator of result
int den = 1; // denominator of result
int fi;
foreach(KeyValuePair it in mp)
{
// if frequency is even
// fi = ci / 2
if (it.Value % 2 == 0)
fi = it.Value / 2;
// if frequency is odd
// fi = ci - 1 / 2.
else
{
fi = (it.Value - 1) / 2;
k++;
}
// sum of all frequencies
num = num + fi;
// product of factorial of
// every frequency
den = den * fact(fi);
}
// if all character are unique
// so there will be no pallindrome,
// so if num != 0 then only we are
// finding the factorial
if (num != 0)
num = fact(num);
int ans = num / den;
if (k != 0)
{
// k are the single
// elements that can be
// placed in middle
ans = ans * k;
}
return (ans);
}
// Driver code
public static void Main(String[] args)
{
String str = "ababab";
int n = str.Length;
Console.WriteLine(numberOfPossiblePallindrome(str, n));
}
}
// This code is contributed by 29AjayKumar 输出：
4
时间复杂度： O(n)