打印矩阵中从左上角到右下角的所有回文路径

给定一个仅包含较低字母字符的矩阵，我们需要打印给定矩阵中的所有回文路径。路径定义为从左上角单元格开始到右下角单元格结束的一系列单元格。我们只能从当前单元格向右和向下移动。我们不能斜着走。
例子：

Input : mat[][] = {"aaab”, 
                   "baaa”
                   “abba”}
Output :aaaaaa, aaaaaa, abaaba

Explanation :
aaaaaa (0, 0) -> (0, 1) -> (1, 1) -> 
       (1, 2) -> (1, 3) -> (2, 3)    
aaaaaa (0, 0) -> (0, 1) -> (0, 2) -> 
       (1, 2) -> (1, 3) -> (2, 3)    
abaaba (0, 0) -> (1, 0) -> (1, 1) -> 
       (1, 2) -> (2, 2) -> (2, 3)

输出数组中元素的顺序无关紧要。

这个想法很简单。我们从左上角 (0, 0) 开始，探索到右下角的所有路径。如果一条路径变成回文，我们打印它。

C++

// C++ program to print all palindromic paths
// from top left to bottom right in a grid.
#include
using namespace std;
#define N 4
bool isPalin(string str)
{
    int len = str.length() / 2;
    for (int i = 0; i < len; i++)
    {
        if (str[i] != str[str.length() - i - 1])
            return false;
    }
    return true;
}
 
// i and j are row and column indexes of current cell
// (initially these are 0 and 0).
void palindromicPath(string str, char a[][N],
                            int i, int j, int m, int n)
{
         
    // If we have not reached bottom right corner, keep
    // exploring
    if (j < m - 1 || i < n - 1)
    {
        if (i < n - 1)
            palindromicPath(str + a[i][j], a, i + 1, j, m, n);
        if (j < m - 1)
            palindromicPath(str + a[i][j], a, i, j + 1, m, n);
    }
 
    // If we reach bottom right corner, we check if
    // if the path used is palindrome or not.
    else {
        str = str + a[n - 1][m - 1];
        if (isPalin(str))
            cout<<(str)<


Java
// Java program to print all palindromic paths
// from top left to bottom right in a grid.
public class PalinPath {
    public static boolean isPalin(String str)
    {
        int len = str.length() / 2;
        for (int i = 0; i < len; i++) {
            if (str.charAt(i) != str.charAt(str.length() - i - 1))
                return false;
        }
        return true;
    }
 
    // i and j are row and column indexes of current cell
    // (initially these are 0 and 0).
    public static void palindromicPath(String str, char a[][],
                                 int i, int j, int m, int n)
    {
           
        // If we have not reached bottom right corner, keep
        // exploring
        if (j < m - 1 || i < n - 1) {
          if (i < n - 1)
             palindromicPath(str + a[i][j], a, i + 1, j, m, n);
         if (j < m - 1)
             palindromicPath(str + a[i][j], a, i, j + 1, m, n);
        }
 
        // If we reach bottom right corner, we check if
        // if the path used is palindrome or not.
        else {
            str = str + a[n - 1][m - 1];
            if (isPalin(str))
                System.out.println(str);
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        char arr[][] = { { 'a', 'a', 'a', 'b' },
                         { 'b', 'a', 'a', 'a' },
                         { 'a', 'b', 'b', 'a' } };
        String str = "";
        palindromicPath(str, arr, 0, 0, 4, 3);
    }
}

Python 3
# Python 3 program to print all
# palindromic paths from top left
# to bottom right in a grid.
 
def isPalin(str):
 
    l = len(str) // 2
    for i in range( l) :
        if (str[i] != str[len(str) - i - 1]):
            return False
         
    return True
 
# i and j are row and column
# indexes of current cell
# (initially these are 0 and 0).
def palindromicPath(str, a, i, j, m, n):
         
    # If we have not reached bottom
    # right corner, keep exploring
    if (j < m - 1 or i < n - 1) :
        if (i < n - 1):
            palindromicPath(str + a[i][j], a,
                            i + 1, j, m, n)
        if (j < m - 1):
            palindromicPath(str + a[i][j], a,
                            i, j + 1, m, n)
 
    # If we reach bottom right corner,
    # we check if the path used is
    # palindrome or not.
    else :
        str = str + a[n - 1][m - 1]
        if isPalin(str):
            print(str)
 
# Driver code
if __name__ == "__main__":
     
    arr = [[ 'a', 'a', 'a', 'b' ],
           ['b', 'a', 'a', 'a' ],
           [ 'a', 'b', 'b', 'a' ]]
    str = ""
    palindromicPath(str, arr, 0, 0, 4, 3)
 
# This code is contributed
# by ChitraNayal

C#
// C# program to print all palindromic paths
// from top left to bottom right in a grid.
using System;
 
class GFG
{
public static bool isPalin(string str)
{
    int len = str.Length / 2;
    for (int i = 0; i < len; i++)
    {
        if (str[i] != str[str.Length - i - 1])
        {
            return false;
        }
    }
    return true;
}
 
// i and j are row and column indexes of
// current cell (initially these are 0 and 0).
public static void palindromicPath(string str, char[][] a,
                                   int i, int j, int m, int n)
{
 
    // If we have not reached bottom
    // right corner, keep exploring
    if (j < m - 1 || i < n - 1)
    {
    if (i < n - 1)
    {
        palindromicPath(str + a[i][j],
                        a, i + 1, j, m, n);
    }
    if (j < m - 1)
    {
        palindromicPath(str + a[i][j],
                        a, i, j + 1, m, n);
    }
    }
 
    // If we reach bottom right corner,
    // we check if the path used is
    // palindrome or not.
    else
    {
        str = str + a[n - 1][m - 1];
        if (isPalin(str))
        {
            Console.WriteLine(str);
        }
    }
}
 
// Driver code
public static void Main(string[] args)
{
    char[][] arr = new char[][]
    {
        new char[] {'a', 'a', 'a', 'b'},
        new char[] {'b', 'a', 'a', 'a'},
        new char[] {'a', 'b', 'b', 'a'}
    };
    string str = "";
    palindromicPath(str, arr, 0, 0, 4, 3);
}
}
 
// This code is contributed by Shrikant13

Javascript

输出：

abaaba
aaaaaa
aaaaaa