给定大小为MxN的矩阵arr [] [] ，任务是查找该矩阵中奇数长度的连续回文序列的数量。
例子：
Input: arr[][] = { { 2, 1, 2 },
{ 1, 1, 1 },
{ 2, 1, 2 }}
Output: 15
Explanation
Contigiuos Palindromic sequences of odd length are:
Row 1: (2), (1), (2), (2, 1, 2) => n(R1) = 4
Row 2: (1), (1), (1), (1, 1, 1) => n(R2) = 4
Row 3: (2), (1), (2), (2, 1, 2) => n(R3) = 4
Column 1: (2, 1, 2) => n(C1) = 1
Column 2: (1, 1, 1) => n(C2) = 1
Column 3: (2, 1, 2) => n(C3) = 1
Therefore,
Total count = n(R1) + n(R2) + n(R3)
+ n(C1) + n(C2) + n(C3)
= 15
Input: arr[][] = { { 1, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 } }
Output: 65
方法：
- 创建一个变量计数以存储矩阵中连续回文序列的总数
- 由于矩阵中的每个元素都是长度为1的连续回文序列，因此将矩阵中元素的总数添加到计数中，即
count += (M*N) - 然后对于长度> 1的序列
- 遍历矩阵的每个元素，并通过将元素与左侧和右侧的其他元素进行比较来计算每行中回文序列的数量
- 类似地，通过将元素与上方和下方的其他元素进行比较，计算每列中回文序列的数量。
- 如果找到，则将找到的回文序列的计数增加1。
- 最后打印计算的回文序列数
下面是上述方法的实现：
C++
// C++ code to Count the odd length contiguous
// Palindromic sequences in the matrix
#include
using namespace std;
#define MAX 10
// Function to count the number of
// contiguous palindromic sequences in the matrix
int countPalindromes(int n, int m, int matrix[MAX][MAX])
{
// Add the total number of elements
// in the matrix to the count
int count = n * m;
// Length of possible sequence to be checked
// for palindrome horizontally and vertically
int length_of_sequence_row;
int length_of_sequence_column;
// Iterate through each element of the matrix
// and count the number of palindromic
// sequences in each row and column
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Find the possible length of sequences
// that can be a palindrome
length_of_sequence_row
= min(j, m - 1 - j);
length_of_sequence_column
= min(i, n - i - 1);
// From i, check if the sequence
// formed by elements to its
// left and right is
// palindrome or not
for (int k = 1; k <= length_of_sequence_row; k++) {
// if the sequence [i, j-k] to [i, j+k]
// is a palindrome,
// increment the count by 1
if (matrix[i][j - k] == matrix[i][j + k]) {
count++;
}
else {
break;
}
}
// From i, check if the sequence
// formed by elements to its
// above and below is
// palindrome or not
for (int k = 1; k <= length_of_sequence_column; k++) {
// if the sequence [i-k, j] to [i+k, j]
// is a palindrome,
// increment the count by 1
if (matrix[i - k][j] == matrix[i + k][j]) {
count++;
}
else {
break;
}
}
}
}
// Return the total count
// of the palindromic sequences
return count;
}
// Driver code
int main(void)
{
int m = 3, n = 3;
int matrix[MAX][MAX] = { { 2, 1, 2 },
{ 1, 1, 1 },
{ 2, 1, 2 } };
cout << countPalindromes(n, m, matrix)
<< endl;
return 0;
} Java
// Java code to Count the odd length contiguous
// Palindromic sequences in the matrix
class GFG
{
static final int MAX = 10;
// Function to count the number of
// contiguous palindromic sequences in the matrix
static int countPalindromes(int n, int m, int matrix[][])
{
// Add the total number of elements
// in the matrix to the count
int count = n * m;
// Length of possible sequence to be checked
// for palindrome horizontally and vertically
int length_of_sequence_row;
int length_of_sequence_column;
// Iterate through each element of the matrix
// and count the number of palindromic
// sequences in each row and column
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Find the possible length of sequences
// that can be a palindrome
length_of_sequence_row
= Math.min(j, m - 1 - j);
length_of_sequence_column
= Math.min(i, n - i - 1);
// From i, check if the sequence
// formed by elements to its
// left and right is
// palindrome or not
for (int k = 1; k <= length_of_sequence_row; k++)
{
// if the sequence [i, j-k] to [i, j+k]
// is a palindrome,
// increment the count by 1
if (matrix[i][j - k] == matrix[i][j + k])
{
count++;
}
else
{
break;
}
}
// From i, check if the sequence
// formed by elements to its
// above and below is
// palindrome or not
for (int k = 1; k <= length_of_sequence_column; k++)
{
// if the sequence [i-k, j] to [i+k, j]
// is a palindrome,
// increment the count by 1
if (matrix[i - k][j] == matrix[i + k][j])
{
count++;
}
else
{
break;
}
}
}
}
// Return the total count
// of the palindromic sequences
return count;
}
// Driver code
public static void main(String []args)
{
int m = 3, n = 3;
int matrix[][] = { { 2, 1, 2 },
{ 1, 1, 1 },
{ 2, 1, 2 } };
System.out.print(countPalindromes(n, m, matrix)
+"\n");
}
}
// This code is contributed by 29AjayKumar
`Python3
# Python code to Count the odd length contiguous
# Palindromic sequences in the matrix
MAX = 10;
# Function to count the number of
# contiguous palindromic sequences in the matrix
def countPalindromes(n, m, matrix):
# Add the total number of elements
# in the matrix to the count
count = n * m;
# Length of possible sequence to be checked
# for palindrome horizontally and vertically
length_of_sequence_row = 0;
length_of_sequence_column = 0;
# Iterate through each element of the matrix
# and count the number of palindromic
# sequences in each row and column
for i in range(n):
for j in range(m):
# Find the possible length of sequences
# that can be a palindrome
length_of_sequence_row = min(j, m - 1 - j);
length_of_sequence_column = min(i, n - i - 1);
# From i, check if the sequence
# formed by elements to its
# left and right is
# palindrome or not
for k in range(1, length_of_sequence_row + 1):
# if the sequence [i, j-k] to [i, j+k]
# is a palindrome,
# increment the count by 1
if (matrix[i][j - k] == matrix[i][j + k]):
count += 1;
else:
break;
# From i, check if the sequence
# formed by elements to its
# above and below is
# palindrome or not
for k in range(1, length_of_sequence_column + 1):
# if the sequence [i-k, j] to [i+k, j]
# is a palindrome,
# increment the count by 1
if (matrix[i - k][j] == matrix[i + k][j]):
count += 1;
else:
break;
# Return the total count
# of the palindromic sequences
return count;
# Driver code
if __name__ == '__main__':
m = 3;
n = 3;
matrix = [ 2, 1, 2 ],[ 1, 1, 1 ],[ 2, 1, 2 ];
print(countPalindromes(n, m, matrix));
# This code is contributed by 29AjayKumarC#
// C# code to Count the odd length contiguous
// Palindromic sequences in the matrix
using System;
class GFG
{
static int MAX = 10;
// Function to count the number of
// contiguous palindromic sequences in the matrix
static int countPalindromes(int n, int m, int [,]matrix)
{
// Add the total number of elements
// in the matrix to the count
int count = n * m;
// Length of possible sequence to be checked
// for palindrome horizontally and vertically
int length_of_sequence_row;
int length_of_sequence_column;
// Iterate through each element of the matrix
// and count the number of palindromic
// sequences in each row and column
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Find the possible length of sequences
// that can be a palindrome
length_of_sequence_row
= Math.Min(j, m - 1 - j);
length_of_sequence_column
= Math.Min(i, n - i - 1);
// From i, check if the sequence
// formed by elements to its
// left and right is
// palindrome or not
for (int k = 1; k <= length_of_sequence_row; k++)
{
// if the sequence [i, j-k] to [i, j+k]
// is a palindrome,
// increment the count by 1
if (matrix[i, j - k] == matrix[i, j + k])
{
count++;
}
else
{
break;
}
}
// From i, check if the sequence
// formed by elements to its
// above and below is
// palindrome or not
for (int k = 1; k <= length_of_sequence_column; k++)
{
// if the sequence [i-k, j] to [i+k, j]
// is a palindrome,
// increment the count by 1
if (matrix[i - k, j] == matrix[i + k, j])
{
count++;
}
else
{
break;
}
}
}
}
// Return the total count
// of the palindromic sequences
return count;
}
// Driver code
public static void Main()
{
int m = 3, n = 3;
int [,]matrix = { { 2, 1, 2 },
{ 1, 1, 1 },
{ 2, 1, 2 } };
Console.WriteLine(countPalindromes(n, m, matrix) );
}
}
// This code is contributed by AnkitRai01输出：
15
时间复杂度： O(n * m * max(n，m))