打印给定字符串的最长回文前缀

给定一个字符串str ，任务是找到给定字符串的最长回文前缀。

例子：

Input: str = “abaac”
Output: aba
Explanation:
The longest prefix of the given string which is palindromic is “aba”.

Input: str = “abacabaxyz”
Output: abacaba
Explanation:
The prefixes of the given string which is palindromic are “aba” and “abacabaxyz”.
But the longest of among two is “abacabaxyz”.

朴素的方法：想法是从起始索引生成给定字符串的所有子串，并检查子串是否是回文的。具有最大长度的回文字符串是结果字符串。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to find the longest prefix
// which is palindromic
void LongestPalindromicPrefix(string s)
{
  
    // Find the length of the given string
    int n = s.length();
  
    // For storing the length of longest
    // prefix palindrome
    int max_len = 0;
  
    // Loop to check the substring of all
    // length from 1 to N which is palindrome
    for (int len = 1; len <= n; len++) {
  
        // String of length i
        string temp = s.substr(0, len);
  
        // To store the reversed of temp
        string temp2 = temp;
  
        // Reversing string temp2
        reverse(temp2.begin(), temp2.end());
  
        // If string temp is palindromic
        // then update the length
        if (temp == temp2) {
            max_len = len;
        }
    }
  
    // Print the palindromic string of
    // max_len
    cout << s.substr(0, max_len);
}
  
// Driver Code
int main()
{
  
    // Given string
    string str = "abaab";
  
    // Function Call
    LongestPalindromicPrefix(str);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to find the longest prefix
// which is palindromic
static void LongestPalindromicPrefix(String s)
{
  
    // Find the length of the given String
    int n = s.length();
  
    // For storing the length of longest
    // prefix palindrome
    int max_len = 0;
  
    // Loop to check the subString of all
    // length from 1 to N which is palindrome
    for (int len = 1; len <= n; len++)
    {
  
        // String of length i
        String temp = s.substring(0, len);
  
        // To store the reversed of temp
        String temp2 = temp;
  
        // Reversing String temp2
        temp2 = reverse(temp2);
  
        // If String temp is palindromic
        // then update the length
        if (temp.equals(temp2))
        {
            max_len = len;
        }
    }
  
    // Print the palindromic String of
    // max_len
    System.out.print(s.substring(0, max_len));
}
  
static String reverse(String input) 
{
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) 
    {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
} 
  
// Driver Code
public static void main(String[] args)
{
  
    // Given String
    String str = "abaab";
  
    // Function Call
    LongestPalindromicPrefix(str);
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach 
  
# Function to find the longest prefix
# which is palindrome
def LongestPalindromicPrefix(string):
      
    # Find the length of the given string
    n = len(string)
      
    # For storing the length of longest 
    # Prefix Palindrome
    max_len = 0
      
    # Loop to check the substring of all 
    # length from 1 to n which is palindrome
    for length in range(0, n + 1):
          
        # String of length i
        temp = string[0:length]
          
        # To store the value of temp
        temp2 = temp
          
        # Reversing the value of temp 
        temp3 = temp2[::-1]
          
        # If string temp is palindromic 
        # then update the length
        if temp == temp3:
            max_len = length
      
    # Print the palindromic string 
    # of max_len
    print(string[0:max_len])
  
# Driver code
if __name__ == '__main__' :
      
    string = "abaac";
      
    # Function call
    LongestPalindromicPrefix(string)
      
# This code is contributed by virusbuddah_

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the longest prefix
// which is palindromic
static void longestPalindromicPrefix(String s)
{
  
    // Find the length of the given String
    int n = s.Length;
  
    // For storing the length of longest
    // prefix palindrome
    int max_len = 0;
  
    // Loop to check the subString of all
    // length from 1 to N which is palindrome
    for (int len = 1; len <= n; len++)
    {
  
        // String of length i
        String temp = s.Substring(0, len);
  
        // To store the reversed of temp
        String temp2 = temp;
  
        // Reversing String temp2
        temp2 = reverse(temp2);
  
        // If String temp is palindromic
        // then update the length
        if (temp.Equals(temp2))
        {
            max_len = len;
        }
    }
  
    // Print the palindromic String of
    // max_len
    Console.Write(s.Substring(0, max_len));
}
  
static String reverse(String input) 
{
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--) 
    {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
} 
  
// Driver Code
public static void Main(String[] args)
{
  
    // Given String
    String str = "abaab";
  
    // Function Call
    longestPalindromicPrefix(str);
}
}
  
// This code is contributed by amal kumar choubey

输出：

aba

时间复杂度： O(N ² ) ，其中 N 是给定字符串的长度。

Efficient Approach：思想是使用预处理算法KMP Algorithm。以下是步骤：

1. 创建一个临时字符串(比如str2 )，它是：

   str2 = str + '?' reverse(str);
   

2. 创建一个字符串str2长度大小的数组(比如lps[] )，它将存储最长的回文前缀，它也是字符串str2的后缀。
3. 使用 KMP 搜索算法的预处理算法更新 lps[]。
4. lps[length(str2) – 1 ] 将给出给定字符串str的最长回文前缀字符串的长度。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to find the longest prefix
// which is palindromic
void LongestPalindromicPrefix(string str)
{
  
    // Create temporary string
    string temp = str + '?';
  
    // Reverse the string str
    reverse(str.begin(), str.end());
  
    // Append string str to temp
    temp += str;
  
    // Find the length of string temp
    int n = temp.length();
  
    // lps[] array for string temp
    int lps[n];
  
    // Intialise every value with zero
    fill(lps, lps + n, 0);
  
    // Iterate the string temp
    for (int i = 1; i < n; i++) {
  
        // Length of longest prefix
        // till less than i
        int len = lps[i - 1];
  
        // Calculate length for i+1
        while (len > 0
               && temp[len] != temp[i]) {
            len = lps[len - 1];
        }
  
        // If character at current index
        // len are same then increament
        // length by 1
        if (temp[i] == temp[len]) {
            len++;
        }
  
        // Update the length at current
        // index to len
        lps[i] = len;
    }
  
    // Print the palindromic string of
    // max_len
    cout << temp.substr(0, lps[n - 1]);
}
  
// Driver's Code
int main()
{
  
    // Given string
    string str = "abaab";
  
    // Function Call
    LongestPalindromicPrefix(str);
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to find the longest 
// prefix which is palindromic
static void LongestPalindromicPrefix(String str)
{
  
    // Create temporary String
    String temp = str + '?';
  
    // Reverse the String str
    str = reverse(str);
  
    // Append String str to temp
    temp += str;
  
    // Find the length of String temp
    int n = temp.length();
  
    // lps[] array for String temp
    int []lps = new int[n];
  
    // Intialise every value with zero
    Arrays.fill(lps, 0);
  
    // Iterate the String temp
    for(int i = 1; i < n; i++)
    {
          
       // Length of longest prefix
       // till less than i
       int len = lps[i - 1];
         
       // Calculate length for i+1
       while (len > 0 && temp.charAt(len) != 
                         temp.charAt(i)) 
       {
           len = lps[len - 1];
       }
         
       // If character at current index
       // len are same then increament
       // length by 1
       if (temp.charAt(i) == temp.charAt(len))
       {
           len++;
       }
         
       // Update the length at current
       // index to len
       lps[i] = len;
    }
  
    // Print the palindromic String 
    // of max_len
    System.out.print(temp.substring(0, lps[n - 1]));
}
  
static String reverse(String input)
{
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
      
    for(l = 0; l < r; l++, r--)
    {
       char temp = a[l];
       a[l] = a[r];
       a[r] = temp;
    }
    return String.valueOf(a);
}
  
// Driver Code
public static void main(String[] args)
{
  
    // Given String
    String str = "abaab";
  
    // Function Call
    LongestPalindromicPrefix(str);
}
}
  
// This code is contributed by Rajput-Ji

蟒蛇3

# Python3 program for the above approach 
  
# Function to find the longest prefix 
# which is palindromic 
def LongestPalindromicPrefix(Str):
  
    # Create temporary string
    temp = Str + "?"
  
    # Reverse the string Str
    Str = Str[::-1]
  
    # Append string Str to temp
    temp = temp + Str
  
    # Find the length of string temp
    n = len(temp)
  
    # lps[] array for string temp
    lps = [0] * n
  
    # Iterate the string temp
    for i in range(1, n):
  
        # Length of longest prefix 
        # till less than i
        Len = lps[i - 1]
  
        # Calculate length for i+1
        while (Len > 0 and temp[Len] != temp[i]):
            Len = lps[Len - 1]
  
        # If character at currrent index
        # Len are same then increment
        # length by 1
        if (temp[i] == temp[Len]):
            Len += 1
  
        # Update the length at current
        # index to Len
        lps[i] = Len
  
    # Print the palindromic string
    # of max_len
    print(temp[0 : lps[n - 1]])
  
# Driver Code
if __name__ == '__main__':
      
    # Given string
    Str = "abaab"
  
    # Fubction call
    LongestPalindromicPrefix(Str)
  
# This code is contributed by himanshu77

C＃

// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the longest 
// prefix which is palindromic
static void longestPalindromicPrefix(String str)
{
      
    // Create temporary String
    String temp = str + '?';
  
    // Reverse the String str
    str = reverse(str);
  
    // Append String str to temp
    temp += str;
  
    // Find the length of String temp
    int n = temp.Length;
  
    // lps[] array for String temp
    int []lps = new int[n];
  
    // Iterate the String temp
    for(int i = 1; i < n; i++)
    {
         
       // Length of longest prefix
       // till less than i
       int len = lps[i - 1];
         
       // Calculate length for i+1
       while (len > 0 && temp[len] != temp[i]) 
       {
           len = lps[len - 1];
       }
         
       // If character at current index
       // len are same then increament
       // length by 1
       if (temp[i] == temp[len])
       {
           len++;
       }
         
       // Update the length at current
       // index to len
       lps[i] = len;
    }
      
    // Print the palindromic String 
    // of max_len
    Console.Write(temp.Substring(0, lps[n - 1]));
}
  
static String reverse(String input)
{
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
      
    for(l = 0; l < r; l++, r--)
    {
       char temp = a[l];
       a[l] = a[r];
       a[r] = temp;
    }
    return String.Join("", a);
}
  
// Driver Code
public static void Main(String[] args)
{
  
    // Given String
    String str = "abaab";
  
    // Function Call
    longestPalindromicPrefix(str);
}
}
  
// This code is contributed by Rajput-Ji

输出：

aba

时间复杂度： O(N) ，其中 N 是给定字符串的长度。
辅助空间： O(N) ，其中 N 是给定字符串的长度。

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