给定一个由小写英文字母组成的字符串str ，任务是找到满足以下条件的最长回文字符串T：
- T = p + m + s其中， p和s分别是给定字符串str的前缀和后缀，字符串m是从字符串删除p和s后的字符串str的前缀或后缀。
- 由p和s串联而成的字符串本身就是回文式。
- 字符串p和s都可以是空字符串。
例子：
Input: str = “abcdfdcecba”
Output: abcdfdcba
Explanation:
Here, p = “abc”
s = “cba”
m = “dfd”
p + s = “abccba” which is a palindrome and m = “dfd” is the prefix after removing the prefix and suffix from the string str. Therefore, T = “abcdfdcba”.
Input: str = “geeksforgeeks”
Output: g
Explanation:
Here, p = “”
s = “g”
m = “”
p + s = “” which is a palindrome and m = “g” is the prefix after removing the prefix and suffix from the string str. Therefore, T = “g”.
方法：针对此问题的想法是将答案分为三部分，以使给定字符串的后缀和前缀的一部分共同组成回文，从而形成答案字符串的开头和结尾。现在，从给定字符串删除这些前缀和后缀之后，我们可以找到回文的最大长度的后缀或前缀字符串(我们可以称其为MidPalindrome)。
因此，答案字符串将由以下形式给出：
answer = prefix + midPalindrome + suffix
可以按照以下步骤计算问题的答案：
- 找到长度，后缀和str的前缀共同构成回文。
- 从str中删除已经形成回文的后缀和前缀子字符串，并将它们存储在单独的字符串。
- 检查其余字符串str中的所有前缀和后缀子字符串，并找到此类字符串最长的一个。
- 最后，将答案的所有三个部分结合起来并返回。
下面是上述方法的实现：
C++
// C++ program to find the longest
// palindrome in a string formed by
// concatenating its prefix and suffix
#include
using namespace std;
// Function to check whether
// the string is a palindrome
bool isPalindrome(string r)
{
string p = r;
// Reverse the string to
// compare with the
// original string
reverse(p.begin(), p.end());
// Check if both are same
return (r == p);
}
// Function to find the longest
// palindrome in a string formed by
// concatenating its prefix and suffix
string PrefixSuffixPalindrome(string str)
{
// Length of the string
int n = str.size(), len = 0;
// Finding the length upto which
// the suffix and prefix forms a
// palindrome together
for (int i = 0; i < n / 2; i++) {
if (str[i] != str[n - i - 1]) {
len = i;
break;
}
}
// Check whether the string
// has prefix and suffix substrings
// which are palindromes.
string prefix = "", suffix = "";
string midPal = "";
// Removing the suffix and prefix
// substrings which already forms
// a palindrome and storing them
// in separate strings
prefix = str.substr(0, len);
suffix = str.substr(n - len);
str = str.substr(len, n - 2 * len);
// Check all prefix substrings
// in the remaining string str
for (int i = 1; i <= str.size(); i++) {
string y = str.substr(0, i);
// Check if the prefix substring
// is a palindrome
if (isPalindrome(y)) {
// If the prefix substring
// is a palindrome then check
// if it is of maximum length
// so far
if (midPal.size() < y.size()) {
midPal = y;
}
}
}
// Check all the suffix substrings
// in the remaining string str
for (int i = 1; i <= str.size(); i++) {
string y = str.substr(str.size() - i);
// Check if the suffix substring
// is a palindrome
if (isPalindrome(y)) {
// If the suffix substring
// is a palindrome then check
// if it is of maximum length
// so far
if (midPal.size() < y.size()) {
midPal = y;
}
}
}
// Combining all the thee parts
// of the answer
string answer = prefix + midPal + suffix;
return answer;
}
// Driver code
int main()
{
string str = "abcdfdcecba";
cout << PrefixSuffixPalindrome(str) << "\n";
return 0;
} Java
// Java program to find the longest
// palindrome in a String formed by
// concatenating its prefix and suffix
import java.util.*;
class GFG{
// Function to check whether
// the String is a palindrome
static boolean isPalindrome(String r)
{
String p = r;
// Reverse the String to
// compare with the
// original String
p = reverse(p);
// Check if both are same
return (r.equals(p));
}
// Function to find the longest
// palindrome in a String formed by
// concatenating its prefix and suffix
static String PrefixSuffixPalindrome(String str)
{
// Length of the String
int n = str.length(), len = 0;
// Finding the length upto which
// the suffix and prefix forms a
// palindrome together
for (int i = 0; i < n / 2; i++) {
if (str.charAt(i) != str.charAt(n - i - 1)) {
len = i;
break;
}
}
// Check whether the String
// has prefix and suffix subStrings
// which are palindromes.
String prefix = "", suffix = "";
String midPal = "";
// Removing the suffix and prefix
// subStrings which already forms
// a palindrome and storing them
// in separate Strings
prefix = str.substring(0, len);
suffix = str.substring(n - len);
str = str.substring(len, (n - 2 * len) + len);
// Check all prefix subStrings
// in the remaining String str
for (int i = 1; i <= str.length(); i++) {
String y = str.substring(0, i);
// Check if the prefix subString
// is a palindrome
if (isPalindrome(y)) {
// If the prefix subString
// is a palindrome then check
// if it is of maximum length
// so far
if (midPal.length() < y.length()) {
midPal = y;
}
}
}
// Check all the suffix subStrings
// in the remaining String str
for (int i = 1; i <= str.length(); i++) {
String y = str.substring(str.length() - i);
// Check if the suffix subString
// is a palindrome
if (isPalindrome(y)) {
// If the suffix subString
// is a palindrome then check
// if it is of maximum length
// so far
if (midPal.length() < y.length()) {
midPal = y;
}
}
}
// Combining all the thee parts
// of the answer
String answer = prefix + midPal + suffix;
return answer;
}
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 str = "abcdfdcecba";
System.out.print(PrefixSuffixPalindrome(str));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to find the longest
# palindrome in a string formed by
# concatenating its prefix and suffix
# Function to check whether
# the string is a palindrome
def isPalindrome(r):
p = r
# Reverse the string to
# compare with the
# original string
p = "".join(reversed(p))
# Check if both are same
return (r == p)
# Function to find the longest
# palindrome in a string formed by
# concatenating its prefix and suffix
def PrefixSuffixPalindrome(st):
# Length of the string
n = len(st)
length = 0
# Finding the length upto which
# the suffix and prefix forms a
# palindrome together
for i in range( n // 2):
if (st[i] != st[n - i - 1]):
length = i
break
# Check whether the string
# has prefix and suffix substrings
# which are palindromes.
prefix = ""
suffix = ""
midPal = ""
# Removing the suffix and prefix
# substrings which already forms
# a palindrome and storing them
# in separate strings
prefix = st[:length]
suffix = st[n - length:]
st = st[length: n - 2 * length+length]
# Check all prefix substrings
# in the remaining string str
for i in range(1,len(st)+1):
y = st[0: i]
# Check if the prefix substring
# is a palindrome
if (isPalindrome(y)):
# If the prefix substring
# is a palindrome then check
# if it is of maximum length
# so far
if (len(midPal) < len(y)):
midPal = y
# Check all the suffix substrings
# in the remaining string str
for i in range(1,len(st)+1):
y = st[len(st)-i]
# Check if the suffix substring
# is a palindrome
if (isPalindrome(y)):
# If the suffix substring
# is a palindrome then check
# if it is of maximum length
# so far
if (len(midPal) < len(y)):
midPal = y
# Combining all the thee parts
# of the answer
answer = prefix + midPal + suffix
return answer
# Driver code
if __name__ == "__main__":
st = "abcdfdcecba";
print(PrefixSuffixPalindrome(st))
# This code is contributed by chitranayalC#
// C# program to find the longest
// palindrome in a String formed by
// concatenating its prefix and suffix
using System;
class GFG{
// Function to check whether
// the String is a palindrome
static bool isPalindrome(String r)
{
String p = r;
// Reverse the String to
// compare with the
// original String
p = reverse(p);
// Check if both are same
return (r.Equals(p));
}
// Function to find the longest
// palindrome in a String formed by
// concatenating its prefix and suffix
static String PrefixSuffixPalindrome(String str)
{
// Length of the String
int n = str.Length, len = 0;
// Finding the length upto which
// the suffix and prefix forms a
// palindrome together
for (int i = 0; i < n / 2; i++) {
if (str[i] != str[n - i - 1]) {
len = i;
break;
}
}
// Check whether the String
// has prefix and suffix subStrings
// which are palindromes.
String prefix = "", suffix = "";
String midPal = "";
// Removing the suffix and prefix
// subStrings which already forms
// a palindrome and storing them
// in separate Strings
prefix = str.Substring(0, len);
suffix = str.Substring(n - len);
str = str.Substring(len, (n - 2 * len) + len);
// Check all prefix subStrings
// in the remaining String str
for (int i = 1; i <= str.Length; i++) {
String y = str.Substring(0, i);
// Check if the prefix subString
// is a palindrome
if (isPalindrome(y)) {
// If the prefix subString
// is a palindrome then check
// if it is of maximum length
// so far
if (midPal.Length < y.Length) {
midPal = y;
}
}
}
// Check all the suffix subStrings
// in the remaining String str
for (int i = 1; i <= str.Length; i++) {
String y = str.Substring(str.Length - i);
// Check if the suffix subString
// is a palindrome
if (isPalindrome(y)) {
// If the suffix subString
// is a palindrome then check
// if it is of maximum length
// so far
if (midPal.Length < y.Length) {
midPal = y;
}
}
}
// Combining all the thee parts
// of the answer
String answer = prefix + midPal + suffix;
return answer;
}
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 str = "abcdfdcecba";
Console.Write(PrefixSuffixPalindrome(str));
}
}
// This code is contributed by 29AjayKumar输出：
abcdfdcba