原文：

给定一个仅包含较低字母字符的矩阵，我们需要在给定的矩阵中打印所有回文路径。 路径定义为从左上角的单元格到右下角的单元格的一系列单元格。 我们只能从当前单元格左右移动。 我们不能斜着走。

例如：

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。 如果路径变成回文，我们将其打印出来。

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 
    // exlporing 
    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 
        // exlporing 
        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 exlporing 
    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 exlporing  
    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

输出：

abaaba
aaaaaa
aaaaaa

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