给定除此以外的小字母字符的字符串str，我们将以索引元组的形式为该字符串提供许多子字符串。我们需要找出给定子字符串范围内回文子字符串的计数。
例子：
Input : String str = "xyaabax"
Range1 = (3, 5)
Range2 = (2, 3)
Output : 4
3
For Range1, substring is "aba"
Count of palindromic substring in "aba" is
four : "a", "b", "aba", "a"
For Range2, substring is "aa"
Count of palindromic substring in "aa" is
3 : "a", "a", "aa"
先决条件：计算字符串中的所有回文子字符串
我们可以使用动态编程解决此问题。首先，我们将制作一个2D数组isPalin，如果字符串(i..j)是回文，则isPalin [i] [j]将为1，否则它将为0。构造isPalin之后，我们将构造另一个2D数组dp dp [i] ] [j]会告诉字符串(i..j)中回文子字符串的计数
现在，我们可以编写isPalin和dp值之间的关系，如下所示，
// isPalin[i][j] will be 1 if ith and jth characters
// are equal and mid substring str(i+1..j-1) is also
// a palindrome
isPalin[i][j] = (str[i] == str[j]) and
(isPalin[i + 1][j – 1])
// Similar to set theory we can write the relation among
// dp values as,
// dp[i][j] will be addition of number of palindromes from
// i to j-1 and i+1 to j subtracting palindromes from i+1
// to j-1 because they are counted twice once in dp[i][j-1]
// and then in dp[i + 1][j] plus 1 if str(i..j) is also a
// palindrome
dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1] +
isPalin[i][j];
用于构造dp数组的解决方案的总时间复杂度为O(length ^ 2)，然后每个查询为O(1)。
C/C++
// C++ program to query number of palindromic
// substrings of a string in a range
#include
using namespace std;
#define M 50
// Utility method to construct the dp array
void constructDP(int dp[M][M], string str)
{
int l = str.length();
// declare 2D array isPalin, isPalin[i][j] will
// be 1 if str(i..j) is palindrome
int isPalin[l + 1][l + 1];
// initialize dp and isPalin array by zeros
for (int i = 0; i <= l; i++) {
for (int j = 0; j <= l; j++) {
isPalin[i][j] = dp[i][j] = 0;
}
}
// loop for starting index of range
for (int i = l - 1; i >= 0; i--) {
// initialize value for one character strings as 1
isPalin[i][i] = 1;
dp[i][i] = 1;
// loop for ending index of range
for (int j = i + 1; j < l; j++) {
/* isPalin[i][j] will be 1 if ith and
jth characters are equal and mid
substring str(i+1..j-1) is also a
palindrome */
isPalin[i][j] = (str[i] == str[j] && (i + 1 > j - 1 || isPalin[i + 1][j - 1]));
/* dp[i][j] will be addition of number
of palindromes from i to j-1 and i+1
to j subtracting palindromes from i+1
to j-1 (as counted twice) plus 1 if
str(i..j) is also a palindrome */
dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + isPalin[i][j];
}
}
}
// method returns count of palindromic substring in range (l, r)
int countOfPalindromeInRange(int dp[M][M], int l, int r)
{
return dp[l][r];
}
// Driver code to test above methods
int main()
{
string str = "xyaabax";
int dp[M][M];
constructDP(dp, str);
int l = 3;
int r = 5;
cout << countOfPalindromeInRange(dp, l, r);
return 0;
} Java
// Java program to query number of palindromic
// substrings of a string in a range
import java.io.*;
class GFG {
// Function to construct the dp array
static void constructDp(int dp[][], String str)
{
int l = str.length();
// declare 2D array isPalin, isPalin[i][j] will
// be 1 if str(i..j) is palindrome
int[][] isPalin = new int[l + 1][l + 1];
// initialize dp and isPalin array by zeros
for (int i = 0; i <= l; i++) {
for (int j = 0; j <= l; j++) {
isPalin[i][j] = dp[i][j] = 0;
}
}
// loop for starting index of range
for (int i = l - 1; i >= 0; i--) {
// initialize value for one character strings as 1
isPalin[i][i] = 1;
dp[i][i] = 1;
// loop for ending index of range
for (int j = i + 1; j < l; j++) {
/* isPalin[i][j] will be 1 if ith and
jth characters are equal and mid
substring str(i+1..j-1) is also a
palindrome */
isPalin[i][j] = (str.charAt(i) == str.charAt(j) && (i + 1 > j - 1 || (isPalin[i + 1][j - 1]) != 0)) ? 1 : 0;
/* dp[i][j] will be addition of number
of palindromes from i to j-1 and i+1
to j subtracting palindromes from i+1
to j-1 (as counted twice) plus 1 if
str(i..j) is also a palindrome */
dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + isPalin[i][j];
}
}
}
// method returns count of palindromic substring in range (l, r)
static int countOfPalindromeInRange(int dp[][], int l, int r)
{
return dp[l][r];
}
// driver program
public static void main(String args[])
{
int MAX = 50;
String str = "xyaabax";
int[][] dp = new int[MAX][MAX];
constructDp(dp, str);
int l = 3;
int r = 5;
System.out.println(countOfPalindromeInRange(dp, l, r));
}
}
// Contributed by Pramod KumarPython3
# Python3 program to query the number of
# palindromic substrings of a string in a range
M = 50
# Utility method to construct the dp array
def constructDP(dp, string):
l = len(string)
# declare 2D array isPalin, isPalin[i][j]
# will be 1 if str(i..j) is palindrome
# and initialize it with zero
isPalin = [[0 for i in range(l + 1)]
for j in range(l + 1)]
# loop for starting index of range
for i in range(l - 1, -1, -1):
# initialize value for one
# character strings as 1
isPalin[i][i], dp[i][i] = 1, 1
# loop for ending index of range
for j in range(i + 1, l):
# isPalin[i][j] will be 1 if ith and jth
# characters are equal and mid substring
# str(i+1..j-1) is also a palindrome
isPalin[i][j] = (string[i] == string[j] and
(i + 1 > j - 1 or isPalin[i + 1][j - 1]))
# dp[i][j] will be addition of number
# of palindromes from i to j-1 and i+1
# to j subtracting palindromes from i+1
# to j-1 (as counted twice) plus 1 if
# str(i..j) is also a palindrome
dp[i][j] = (dp[i][j - 1] + dp[i + 1][j] -
dp[i + 1][j - 1] + isPalin[i][j])
# Method returns count of palindromic
# substring in range (l, r)
def countOfPalindromeInRange(dp, l, r):
return dp[l][r]
# Driver code
if __name__ == "__main__":
string = "xyaabax"
dp = [[0 for i in range(M)]
for j in range(M)]
constructDP(dp, string)
l, r = 3, 5
print(countOfPalindromeInRange(dp, l, r))
# This code is contributed by Rituraj JainC#
// C# program to query number of palindromic
// substrings of a string in a range
using System;
class GFG {
// Function to construct the dp array
static void constructDp(int[, ] dp, string str)
{
int l = str.Length;
// declare 2D array isPalin, isPalin[i][j]
// will be 1 if str(i..j) is palindrome
int[, ] isPalin = new int[l + 1, l + 1];
// initialize dp and isPalin array by zeros
for (int i = 0; i <= l; i++) {
for (int j = 0; j <= l; j++) {
isPalin[i, j] = dp[i, j] = 0;
}
}
// loop for starting index of range
for (int i = l - 1; i >= 0; i--) {
// initialize value for one
// character strings as 1
isPalin[i, i] = 1;
dp[i, i] = 1;
// loop for ending index of range
for (int j = i + 1; j < l; j++) {
/* isPalin[i][j] will be 1 if ith and
jth characters are equal and mid
substring str(i+1..j-1) is also a
palindrome*/
isPalin[i, j] = (str[i] == str[j] && (i + 1 > j - 1 ||
(isPalin[i + 1, j - 1]) != 0)) ? 1 : 0;
/* dp[i][j] will be addition of number
of palindromes from i to j-1 and i+1
to j subtracting palindromes from i+1
to j-1 (as counted twice) plus 1 if
str(i..j) is also a palindrome */
dp[i, j] = dp[i, j - 1] + dp[i + 1, j] -
dp[i + 1, j - 1] + isPalin[i, j];
}
}
}
// method returns count of palindromic
// substring in range (l, r)
static int countOfPalindromeInRange(int[, ] dp,
int l, int r)
{
return dp[l, r];
}
// driver program
public static void Main()
{
int MAX = 50;
string str = "xyaabax";
int[, ] dp = new int[MAX, MAX];
constructDp(dp, str);
int l = 3;
int r = 5;
Console.WriteLine(countOfPalindromeInRange(dp, l, r));
}
}
// This code is contributed by vt_m.PHP
= 0; $i--)
{
// initialize value for one character
// strings as 1
$isPalin[$i][$i] = 1;
$dp[$i][$i] = 1;
// loop for ending index of range
for ($j = $i + 1; $j < $l; $j++)
{
/* isPalin[i][j] will be 1 if ith and
jth characters are equal and mid
substring str(i+1..j-1) is also a
palindrome */
$isPalin[$i][$j] = ($str[$i] == $str[$j] &&
($i + 1 > $j - 1 ||
$isPalin[$i + 1][$j - 1]));
/* dp[i][j] will be addition of number
of palindromes from i to j-1 and i+1
to j subtracting palindromes from i+1
to j-1 (as counted twice) plus 1 if
str(i..j) is also a palindrome */
$dp[$i][$j] = $dp[$i][$j - 1] + $dp[$i + 1][$j] -
$dp[$i + 1][$j - 1] + $isPalin[$i][$j];
}
}
return $dp ;
}
// method returns count of palindromic
// substring in range (l, r)
function countOfPalindromeInRange($dp, $l, $r)
{
return $dp[$l][$r];
}
// Driver code
$str = "xyaabax";
$dp = array(array());
for($i = 0; $i < $GLOBALS['M']; $i++ )
for($j = 0; $j < $GLOBALS['M']; $j++)
$dp[$i][$j] = 0;
$dp = constructDP($dp, $str);
$l = 3;
$r = 5;
echo countOfPalindromeInRange($dp, $l, $r);
// This code is contributed by Ryuga
?>输出：
4