从词法上讲，给定字符串的所有最短回文子字符串

给定大小为N的字符串s 。任务是从字典上查找给定字符串所有最短的回文子字符串。

例子：

Input: s= “programming”
Output: a g i m n o p r
Explanation:
The Lexicographical shortest palindrome substring for the word “programming” will be the single characters from the given string. Hence, the output is : a g i m n o p r.

Input: s= “geeksforgeeks”
Output: e f g k o r s

方法：

为了解决上述问题，最先观察到的是，最短回文子字符串的大小为1。因此，根据问题陈述，我们必须按字典顺序找到大小为1的所有不同子字符串，这意味着给定字符串中的所有字符。

下面是上述方法的实现：

C++

// C++ program to find Lexicographically all
// Shortest Palindromic Substrings from a given string
  
#include 
using namespace std;
  
// Function to find all lexicographically
// shortest palindromic substring
void shortestPalindrome(string s)
{
  
    // Array to keep track of alphabetic characters
    int abcd[26] = { 0 };
  
    for (int i = 0; i < s.length(); i++)
        abcd[s[i] - 97] = 1;
  
    // Iterate to print all lexicographically shortest substring
    for (int i = 0; i < 26; i++) {
        if (abcd[i] == 1)
            cout << char(i + 97) << " ";
    }
}
  
// Driver code
int main()
{
    string s = "geeksforgeeks";
  
    shortestPalindrome(s);
  
    return 0;
}

Java

// Java program to find Lexicographically all
// Shortest Palindromic Substrings from a given string
class Main
{
    // Function to find all lexicographically
    // shortest palindromic substring
    static void shortestPalindrome(String s)
    {
  
        // Array to keep track of 
        // alphabetic characters
        int[] abcd = new int[26];
  
        for (int i = 0; i < s.length(); i++)
            abcd[s.charAt(i) - 97] = 1;
  
        // Iterate to print all lexicographically
        // shortest substring
        for (int i = 0; i < 26; i++)
        {
            if (abcd[i] == 1) 
            {
                System.out.print((char)(i + 97) + " ");
            }
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String s = "geeksforgeeks";
        shortestPalindrome(s);
    }
}

Python3

# C++ program to find Lexicographically all
# Shortest Palindromic Substrings from a given string
  
# Function to find all lexicographically 
# shortest palindromic substring
def shortestPalindrome (s) :
      
    # Array to keep track of alphabetic characters
    abcd = [0]*26
  
    for i in range(len(s)):
        abcd[ord(s[i])-97] = 1
      
    # Iterate to print all lexicographically shortest substring
    for i in range(26): 
        if abcd[i]== 1 :
            print( chr(i + 97), end =' ' )
  
# Driver code
s = "geeksforgeeks"
  
shortestPalindrome (s)

C#

// C# program to find Lexicographically 
// all shortest palindromic substrings
// from a given string
using System;
  
class GFG{
      
// Function to find all lexicographically 
// shortest palindromic substring 
static void shortestPalindrome(string s) 
{ 
  
    // Array to keep track of
    // alphabetic characters 
    int[] abcd = new int[26]; 
  
    for(int i = 0; i < s.Length; i++) 
       abcd[s[i] - 97] = 1; 
  
    // Iterate to print all lexicographically 
    // shortest substring 
    for(int i = 0; i < 26; i++)
    { 
       if (abcd[i] == 1)
       { 
           Console.Write((char)(i + 97) + " "); 
       } 
    } 
} 
  
// Driver code 
static public void Main(string[] args) 
{ 
    string s = "geeksforgeeks"; 
    shortestPalindrome(s); 
} 
} 
  
// This code is contributed by AnkitRai01

输出：

e f g k o r s

时间复杂度： O(N)，其中N是字符串的大小。

空间复杂度： O(1)