给定字符串str ，任务是将字符串分成最小部分，以使每个部分具有相同的长度，并且每个部分都是回文。打印所需数量的零件。
例子：
Input: str = “civicbob”
Output: 8
“b”, “b”, “c”, “c”, “i”, “i”, “v” and “o” are the required partitions. “civic” and “bob” are also palindromes but they are not of equal length
Input: str = “noonpeep”
Output: 1
方法：
- 计算出现的奇数字符并将其存储在countOdd中。
- 对所有偶数出现的字符的频率求和，并将其存储在sumEven中。
- 由于同一回文的一部分不能超过一个奇数频率字符。因此，如果(sumEven / 2)％countOdd = 0，则答案为countOdd，因为sumEven可分为countOdd部分。
- 另外，我们仍然可以将出现的奇数字符分成回文对。例如，如果字符a出现3次，即aaa，则aa可以是某些回文集的一部分，而保留a为奇数(不影响原始频率)。
下面是上述方法的实现：
CPP
// C++ implementation of the approach
#include
using namespace std;
// Function to return the frequency array
// for the given string
int* getFrequencies(string str)
{
static int freq[26] = { 0 };
for (int i = 0; i < str.length(); i++) {
freq[str[i] - 'a']++;
}
return freq;
}
// Function to return the required count
int countMinParts(string str)
{
int n = str.length();
int *freq = getFrequencies(str);
vector oddFreq, evenFreq;
int i, sumEven = 0;
for (i = 0; i < 26; i++) {
if (freq[i] == 0)
continue;
// Add frequencies of the even appearing
// characters
if (freq[i] % 2 == 0)
evenFreq.push_back(freq[i]);
// Count of the characters that appeared
// odd number of times
else
oddFreq.push_back(freq[i]);
}
for (i = 0; i < evenFreq.size(); i++) {
sumEven += evenFreq[i];
}
// If there are no characters with odd frequency
if (oddFreq.size() == 0)
return 1;
// If there are no characters with even frequency
if (sumEven == 0) {
// Only a single character with odd frequency
if (oddFreq.size() == 1)
return 1;
// More than 1 character with odd frequency
// string isn't a palindrome
return 0;
}
i = 0;
// All odd appearing characters can also contribute to
// the even length palindrome if one character
// is removed from the frequency leaving it as even
while (i < oddFreq.size()) {
// If k palindromes are possible where k
// is the number of characters with odd frequency
if ((sumEven / 2) % oddFreq.size() == 0)
return oddFreq.size();
// Current character can no longer be an element
// in a string other than the mid character
if (oddFreq[i] == 1) {
i++;
continue;
}
// If current character has odd frequency > 1
// take two characters which can be used in
// any of the parts
sumEven += 2;
// Update the frequency
oddFreq[i] = oddFreq[i] - 2;
}
// If not possible, then every character of the
// string will act as a separate palindrome
return n;
}
// Driver code
int main()
{
string s = "noonpeep";
cout< Java
// Java implementation of the approach
import java.util.*;
public class GFG {
// Function to return the frequency array
// for the given string
static int[] getFrequencies(String str)
{
int freq[] = new int[26];
for (int i = 0; i < str.length(); i++) {
freq[str.charAt(i) - 'a']++;
}
return freq;
}
// Function to return the required count
static int countMinParts(String str)
{
int n = str.length();
int freq[] = getFrequencies(str);
List oddFreq = new ArrayList<>();
List evenFreq = new ArrayList<>();
int i, sumEven = 0;
for (i = 0; i < 26; i++) {
if (freq[i] == 0)
continue;
// Add frequencies of the even appearing
// characters
if (freq[i] % 2 == 0)
evenFreq.add(freq[i]);
// Count of the characters that appeared
// odd number of times
else
oddFreq.add(freq[i]);
}
for (i = 0; i < evenFreq.size(); i++) {
sumEven += evenFreq.get(i);
}
// If there are no characters with odd frequency
if (oddFreq.size() == 0)
return 1;
// If there are no characters with even frequency
if (sumEven == 0) {
// Only a single character with odd frequency
if (oddFreq.size() == 1)
return 1;
// More than 1 character with odd frequency
// string isn't a palindrome
return 0;
}
i = 0;
// All odd appearing characters can also contribute to
// the even length palindrome if one character
// is removed from the frequency leaving it as even
while (i < oddFreq.size()) {
// If k palindromes are possible where k
// is the number of characters with odd frequency
if ((sumEven / 2) % oddFreq.size() == 0)
return oddFreq.size();
// Current character can no longer be an element
// in a string other than the mid character
if (oddFreq.get(i) == 1) {
i++;
continue;
}
// If current character has odd frequency > 1
// take two characters which can be used in
// any of the parts
sumEven += 2;
// Update the frequency
oddFreq.set(i, oddFreq.get(i) - 2);
}
// If not possible, then every character of the
// string will act as a separate palindrome
return n;
}
// Driver code
public static void main(String[] args)
{
String s = "noonpeep";
System.out.println(countMinParts(s));
}
}
// This code is contributed by chitranayal Python3
# Python3 implementation of the approach
# Function to return the frequency array
# for the given string
def getFrequencies(string) :
freq = [0] * 26
for i in range(len(string)) :
freq[ord(string[i]) -
ord('a')] += 1
return freq
# Function to return the required count
def countMinParts(string) :
n = len(string)
freq = getFrequencies(string)
oddFreq = []
evenFreq = []
sumEven = 0
for i in range(26) :
if freq[i] == 0 :
continue
# Add frequencies of the even
# appearing characters
if freq[i] % 2 == 0 :
evenFreq.append(freq[i])
# Count of the characters that
# appeared odd number of times
else :
oddFreq.append(freq[i])
for i in range(len(evenFreq)) :
sumEven += evenFreq[i]
# If there are no characters with
# odd frequency
if len(oddFreq) == 0 :
return 1
# If there are no characters with
# even frequency
if sumEven == 0 :
# Only a single character with
# odd frequency
if len(oddFreq) == 1:
return 1
# More than 1 character with odd
# frequency string isn't a palindrome
return 0
i = 0
# All odd appearing characters can also
# contribute to the even length palindrome
# if one character is removed from the
# frequency leaving it as even
while(i < len(oddFreq)) :
# If k palindromes are possible where
# k is the number of characters with
# odd frequency
if ((sumEven / 2) % len(oddFreq) == 0) :
return len(oddFreq)
# Current character can no longer be
# an element in a string other than
# the mid character
if (oddFreq[i] == 1) :
i += 1
continue
# If current character has odd frequency > 1
# take two characters which can be used in
# any of the parts
sumEven += 2
# Update the frequency
oddFreq[i] = oddFreq[i] - 2
# If not possible, then every character of the
# string will act as a separate palindrome
return n
# Driver code
if __name__ == "__main__" :
s = "noonpeep"
print(countMinParts(s))
# This code is contributed by RyugaC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the frequency array
// for the given string
static int[] getFrequencies(String str)
{
int []freq = new int[26];
for (int i = 0; i < str.Length; i++)
{
freq[str[i] - 'a']++;
}
return freq;
}
// Function to return the required count
static int countMinParts(String str)
{
int n = str.Length;
int []freq = getFrequencies(str);
List oddFreq = new List();
List evenFreq = new List();
int i, sumEven = 0;
for (i = 0; i < 26; i++)
{
if (freq[i] == 0)
continue;
// Add frequencies of the even appearing
// characters
if (freq[i] % 2 == 0)
evenFreq.Add(freq[i]);
// Count of the characters that appeared
// odd number of times
else
oddFreq.Add(freq[i]);
}
for (i = 0; i < evenFreq.Count; i++)
{
sumEven += evenFreq[i];
}
// If there are no characters with odd frequency
if (oddFreq.Count == 0)
return 1;
// If there are no characters with even frequency
if (sumEven == 0)
{
// Only a single character with odd frequency
if (oddFreq.Count == 1)
return 1;
// More than 1 character with odd frequency
// string isn't a palindrome
return 0;
}
i = 0;
// All odd appearing characters can also contribute to
// the even length palindrome if one character
// is removed from the frequency leaving it as even
while (i < oddFreq.Count)
{
// If k palindromes are possible where k
// is the number of characters with odd frequency
if ((sumEven / 2) % oddFreq.Count == 0)
return oddFreq.Count;
// Current character can no longer be an element
// in a string other than the mid character
if (oddFreq[i] == 1)
{
i++;
continue;
}
// If current character has odd frequency > 1
// take two characters which can be used in
// any of the parts
sumEven += 2;
// Update the frequency
oddFreq.Insert(i, oddFreq[i] - 2);
}
// If not possible, then every character of the
// string will act as a separate palindrome
return n;
}
// Driver code
public static void Main(String[] args)
{
String s = "noonpeep";
Console.WriteLine(countMinParts(s));
}
}
// This code has been contributed by 29AjayKumar 输出：
1
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