从给定字符串
给定一个字符串S ，任务是将S的字符划分为最少数量的回文字符串。
注意：可以有多个正确答案。
例子：
Input: S = “geeksforgeeks”
Output: {eegksrskgee, o, f}
Explanation:
There should be at least 3 strings “eegksrskgee”, “o”, “f”. All 3 formed strings are palindrome.
Input: S = “abbaa”
Output: {“ababa”}
Explanation:
The given string S = “ababa” is already a palindrome string so, only 1 string will be formed.
Input: S = “abc”
Output: {“a”, “b”, “c”}
Explanation:
There should be at least 3 strings “a”, “b”, “c”. All 3 formed strings are palindrome.
方法：
主要观察结果是，如果反转回文字符串，则保持不变。形成的字符串的长度无关紧要，因此尽量使字符串尽可能大。如果某个字符不能成为回文字符串的一部分，则将该字符推入一个新字符串。
- 一般的方法是存储每个字符的频率，如果某些字符的频率是奇数，那么我们必须创建一个新字符串并将我们的答案增加 1。
- 请注意，如果没有奇频字符，则至少还需要一个字符串，因此最终答案将是 max(1,奇频字符)。
- 要打印字符串，将奇数字符频率字符存储在一个向量中，将偶数字符存储在另一个向量中。
- 用奇数个字符字符串一个回文字符串，并附加一个偶数个字符的字符串，所有其他字符串都只有一个字符。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to return
// minimum palindromic strings formed
int minimumPalindromicStrings(string S)
{
// Store the length of string
int N = S.length();
// Iterate over the string
// and store the frequency of
// all characters A vector
// to store the frequency
vector freq(26);
for (int i = 0; i < N; i++) {
freq[S[i] - 'a']++;
}
// Store the odd frequency characters
vector oddFreqCharacters;
// Store the even frequency of characters
// in the same vector by dividing
// current frequency by 2 as one element
// will be used on left as well as right side
// Iterate through all possible characters
// and check the parity of frequency
for (int i = 0; i < 26; i++) {
if (freq[i] & 1) {
oddFreqCharacters.push_back(i);
freq[i]--;
}
freq[i] /= 2;
}
// store answer in a vector
vector ans;
// if there are no odd Frequency characters
// then print the string with even characters
if (oddFreqCharacters.empty()) {
// store the left part
// of the palindromic string
string left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += char(i + 'a');
}
}
string right = left;
// reverse the right part of the string
reverse(right.begin(), right.end());
// store the string in ans vector
ans.push_back(left + right);
}
else {
// take any character
// from off frequency element
string middle = "";
int c = oddFreqCharacters.back();
oddFreqCharacters.pop_back();
middle += char(c + 'a');
// repeat the above step to form
// left and right strings
string left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += char(i + 'a');
}
}
string right = left;
// reverse the right part of the string
reverse(right.begin(), right.end());
// store the string in ans vector
ans.push_back(left + middle + right);
// store all other odd frequency strings
while (!oddFreqCharacters.empty()) {
int c = oddFreqCharacters.back();
oddFreqCharacters.pop_back();
string middle = "";
middle += char(c + 'a');
ans.push_back(middle);
}
}
// Print the answer
cout << "[";
for (int i = 0; i < ans.size(); i++) {
cout << ans[i];
if (i != ans.size() - 1) {
cout << ", ";
}
}
cout << "]";
return 0;
}
// Driver Code
int main()
{
string S = "geeksforgeeks";
minimumPalindromicStrings(S);
} Java
// Java Program to implement
// the above approach
import java.util.*;
class GFG{
// Function to return
// minimum palindromic Strings formed
static int minimumPalindromicStrings(String S)
{
// Store the length of String
int N = S.length();
// Iterate over the String
// and store the frequency of
// all characters A vector
// to store the frequency
int[] freq = new int[26];
for (int i = 0; i < N; i++) {
freq[S.charAt(i) - 'a']++;
}
// Store the odd frequency characters
Vector oddFreqCharacters = new Vector();
// Store the even frequency of characters
// in the same vector by dividing
// current frequency by 2 as one element
// will be used on left as well as right side
// Iterate through all possible characters
// and check the parity of frequency
for (int i = 0; i < 26; i++) {
if (freq[i] % 2 == 1) {
oddFreqCharacters.add(i);
freq[i]--;
}
freq[i] /= 2;
}
// store answer in a vector
Vector ans = new Vector();
// if there are no odd Frequency characters
// then print the String with even characters
if (oddFreqCharacters.isEmpty()) {
// store the left part
// of the palindromic String
String left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += (char)(i + 'a');
}
}
String right = left;
// reverse the right part of the String
right = reverse(right);
// store the String in ans vector
ans.add(left + right);
}
else {
// take any character
// from off frequency element
String middle = "";
int c = oddFreqCharacters.lastElement();
oddFreqCharacters.remove(oddFreqCharacters.size()-1);
middle += (char)(c + 'a');
// repeat the above step to form
// left and right Strings
String left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += (char)(i + 'a');
}
}
String right = left;
// reverse the right part of the String
right = reverse(right);
// store the String in ans vector
ans.add(left + middle + right);
// store all other odd frequency Strings
while (!oddFreqCharacters.isEmpty()) {
c = oddFreqCharacters.lastElement();
oddFreqCharacters.remove(oddFreqCharacters.size()-1);
middle = "";
middle += (char)(c + 'a');
ans.add(middle);
}
}
// Print the answer
System.out.print("[");
for (int i = 0; i < ans.size(); i++) {
System.out.print(ans.get(i));
if (i != ans.size() - 1) {
System.out.print(", ");
}
}
System.out.print("]");
return 0;
}
static String reverse(String input) {
char[] a = input.toCharArray();
int l, r = a.length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
// Driver Code
public static void main(String[] args)
{
String S = "geeksforgeeks";
minimumPalindromicStrings(S);
}
}
// This code is contributed by 29AjayKumar C#
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
public class GFG{
// Function to return
// minimum palindromic Strings formed
static int minimumPalindromicStrings(String S)
{
// Store the length of String
int N = S.Length;
// Iterate over the String
// and store the frequency of
// all characters A vector
// to store the frequency
int[] freq = new int[26];
for (int i = 0; i < N; i++) {
freq[S[i] - 'a']++;
}
// Store the odd frequency characters
List oddFreqchars = new List();
// Store the even frequency of characters
// in the same vector by dividing
// current frequency by 2 as one element
// will be used on left as well as right side
// Iterate through all possible characters
// and check the parity of frequency
for (int i = 0; i < 26; i++) {
if (freq[i] % 2 == 1) {
oddFreqchars.Add(i);
freq[i]--;
}
freq[i] /= 2;
}
// store answer in a vector
List ans = new List();
// if there are no odd Frequency characters
// then print the String with even characters
if (oddFreqchars.Count==0) {
// store the left part
// of the palindromic String
String left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += (char)(i + 'a');
}
}
String right = left;
// reverse the right part of the String
right = reverse(right);
// store the String in ans vector
ans.Add(left + right);
}
else {
// take any character
// from off frequency element
String middle = "";
int c = oddFreqchars[oddFreqchars.Count-1];
oddFreqchars.RemoveAt(oddFreqchars.Count-1);
middle += (char)(c + 'a');
// repeat the above step to form
// left and right Strings
String left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += (char)(i + 'a');
}
}
String right = left;
// reverse the right part of the String
right = reverse(right);
// store the String in ans vector
ans.Add(left + middle + right);
// store all other odd frequency Strings
while (oddFreqchars.Count!=0) {
c = oddFreqchars[oddFreqchars.Count-1];
oddFreqchars.RemoveAt(oddFreqchars.Count-1);
middle = "";
middle += (char)(c + 'a');
ans.Add(middle);
}
}
// Print the answer
Console.Write("[");
for (int i = 0; i < ans.Count; i++) {
Console.Write(ans[i]);
if (i != ans.Count - 1) {
Console.Write(", ");
}
}
Console.Write("]");
return 0;
}
static String reverse(String input) {
char[] a = input.ToCharArray();
int l, r = a.Length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("",a);
}
// Driver Code
public static void Main(String[] args)
{
String S = "geeksforgeeks";
minimumPalindromicStrings(S);
}
}
// This code is contributed by 29AjayKumar 输出：
[eegksrskgee, o, f]时间复杂度： O(N*26)
空间复杂度： O(N)