给定一个字符串str和一个字符ch ，任务是找到str的最长回文子字符串，使其以给定字符ch开始和结束。
例子：
Input: str = “lapqooqpqpl”, ch = ‘p’
Output: 6
“pqooqp” is the maximum length palindromic
sub-string that starts and ends with ‘p’.
Input: str = “geeksforgeeks”, ch = ‘k’
Output: 1
“k” is the valid sub-string.
方法：对于每个可能的索引对(i，j) ，以使str [i] = str [j] = ch检查子字符串str [i…j]是否为回文。对于所有发现的回文，请存储迄今为止找到的最长回文的长度。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns true if
// str[i...j] is a palindrome
bool isPalindrome(string str, int i, int j)
{
while (i < j) {
if (str[i] != str[j])
return false;
i++;
j--;
}
return true;
}
// Function to return the length of the
// longest palindromic sub-string such that
// it starts and ends with the character ch
int maxLenPalindrome(string str, int n, char ch)
{
int maxLen = 0;
for (int i = 0; i < n; i++) {
// If current character is
// a valid starting index
if (str[i] == ch) {
// Instead of finding the ending index from
// the beginning, find the index from the end
// This is because if the current sub-string
// is a palindrome then there is no need to check
// the sub-strings of smaller length and we can
// skip to the next iteration of the outer loop
for (int j = n - 1; j >= i; j--) {
// If current character is
// a valid ending index
if (str[j] == ch) {
// If str[i...j] is a palindrome then update
// the length of the maximum palindrome so far
if (isPalindrome(str, i, j)) {
maxLen = max(maxLen, j - i + 1);
break;
}
}
}
}
}
return maxLen;
}
// Driver code
int main()
{
string str = "lapqooqpqpl";
int n = str.length();
char ch = 'p';
cout << maxLenPalindrome(str, n, ch);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function that returns true if
// str[i...j] is a palindrome
static boolean isPalindrome(String str,
int i, int j)
{
while (i < j)
{
if (str.charAt(i) != str.charAt(j))
{
return false;
}
i++;
j--;
}
return true;
}
// Function to return the length of the
// longest palindromic sub-string such that
// it starts and ends with the character ch
static int maxLenPalindrome(String str, int n, char ch)
{
int maxLen = 0;
for (int i = 0; i < n; i++)
{
// If current character is
// a valid starting index
if (str.charAt(i) == ch)
{
// Instead of finding the ending index from
// the beginning, find the index from the end
// This is because if the current sub-string
// is a palindrome then there is no need to check
// the sub-strings of smaller length and we can
// skip to the next iteration of the outer loop
for (int j = n - 1; j >= i; j--)
{
// If current character is
// a valid ending index
if (str.charAt(j) == ch)
{
// If str[i...j] is a palindrome then update
// the length of the maximum palindrome so far
if (isPalindrome(str, i, j))
{
maxLen = Math.max(maxLen, j - i + 1);
break;
}
}
}
}
}
return maxLen;
}
// Driver code
public static void main(String[] args)
{
String str = "lapqooqpqpl";
int n = str.length();
char ch = 'p';
System.out.println(maxLenPalindrome(str, n, ch));
}
}
// This code is contributed by Princi SinghPython3
# Python implementation of the approach
# Function that returns true if
# str[i...j] is a palindrome
def isPalindrome(str, i, j):
while (i < j):
if (str[i] != str[j]):
return False;
i+=1;
j-=1;
return True;
# Function to return the length of the
# longest palindromic sub-string such that
# it starts and ends with the character ch
def maxLenPalindrome(str, n, ch):
maxLen = 0;
for i in range(n):
# If current character is
# a valid starting index
if (str[i] == ch):
# Instead of finding the ending index from
# the beginning, find the index from the end
# This is because if the current sub-string
# is a palindrome then there is no need to check
# the sub-strings of smaller length and we can
# skip to the next iteration of the outer loop
for j in range(n-1,i+1,-1):
# If current character is
# a valid ending index
if (str[j] == ch):
# If str[i...j] is a palindrome then update
# the length of the maximum palindrome so far
if (isPalindrome(str, i, j)):
maxLen = max(maxLen, j - i + 1);
break;
return maxLen;
# Driver code
str = "lapqooqpqpl";
n = len(str);
ch = 'p';
print(maxLenPalindrome(str, n, ch));
# This code is contributed by 29AjayKumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if
// str[i...j] is a palindrome
static bool isPalindrome(string str,
int i, int j)
{
while (i < j)
{
if (str[i] != str[j])
{
return false;
}
i++;
j--;
}
return true;
}
// Function to return the length of the
// longest palindromic sub-string such that
// it starts and ends with the character ch
static int maxLenPalindrome(string str, int n, char ch)
{
int maxLen = 0;
for (int i = 0; i < n; i++)
{
// If current character is
// a valid starting index
if (str[i] == ch)
{
// Instead of finding the ending index from
// the beginning, find the index from the end
// This is because if the current sub-string
// is a palindrome then there is no need to check
// the sub-strings of smaller length and we can
// skip to the next iteration of the outer loop
for (int j = n - 1; j >= i; j--)
{
// If current character is
// a valid ending index
if (str[j] == ch)
{
// If str[i...j] is a palindrome then update
// the length of the maximum palindrome so far
if (isPalindrome(str, i, j))
{
maxLen = Math.Max(maxLen, j - i + 1);
break;
}
}
}
}
}
return maxLen;
}
// Driver code
public static void Main()
{
string str = "lapqooqpqpl";
int n = str.Length;
char ch = 'p';
Console.WriteLine(maxLenPalindrome(str, n, ch));
}
}
// This code is contributed by AnkitRai01输出：
6