最长回文子串的长度

给定一个长度为N的字符串S ，任务是从给定的字符串找出最长回文子串的长度。

例子：

Input: S = “abcbab”
Output: 5
Explanation: 
string “abcba” is the longest substring that is a palindrome which is of length 5.

Input: S = “abcdaa”
Output: 2
Explanation: 
string “aa” is the longest substring that is a palindrome which is of length 2. 

朴素方法：解决问题的最简单方法是生成给定字符串的所有可能子串，并打印最长子串的长度，即回文。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to obtain the length of
// the longest palindromic substring
int longestPalSubstr(string str)
{
    // Length of given string
    int n = str.size();
 
    // Stores the maximum length
    int maxLength = 1, start = 0;
 
    // Iterate over the string
    for (int i = 0;
         i < str.length(); i++) {
 
        // Iterate over the string
        for (int j = i;
             j < str.length(); j++) {
            int flag = 1;
 
            // Check for palindrome
            for (int k = 0;
                 k < (j - i + 1) / 2; k++)
                if (str[i + k]
                    != str[j - k])
                    flag = 0;
 
            // If string [i, j - i + 1]
            // is palindromic
            if (flag
                && (j - i + 1) > maxLength) {
                start = i;
                maxLength = j - i + 1;
            }
        }
    }
 
    // Return length of LPS
    return maxLength;
}
 
// Driver Code
int main()
{
    // Given string
    string str = "forgeeksskeegfor";
 
    // Function Call
    cout << longestPalSubstr(str);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
  
// Function to obtain the length of
// the longest palindromic substring
static int longestPalSubstr(String str)
{
     
    // Length of given string
    int n = str.length();
  
    // Stores the maximum length
    int maxLength = 1, start = 0;
  
    // Iterate over the string
    for(int i = 0; i < str.length(); i++)
    {
         
        // Iterate over the string
        for(int j = i; j < str.length(); j++)
        {
            int flag = 1;
  
            // Check for palindrome
            for(int k = 0;
                    k < (j - i + 1) / 2; k++)
                if (str.charAt(i + k) !=
                    str.charAt(j - k))
                    flag = 0;
  
            // If string [i, j - i + 1]
            // is palindromic
            if (flag != 0 &&
               (j - i + 1) > maxLength)
            {
                start = i;
                maxLength = j - i + 1;
            }
        }
    }
  
    // Return length of LPS
    return maxLength;
}
  
// Driver Code
public static void main (String[] args)
{
     
    // Given string
    String str = "forgeeksskeegfor";
  
    // Function call
    System.out.print(longestPalSubstr(str));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
 
# Function to obtain the length of
# the longest palindromic substring
def longestPalSubstr(str):
     
    # Length of given string
    n = len(str)
  
    # Stores the maximum length
    maxLength = 1
    start = 0
  
    # Iterate over the string
    for i in range(len(str)):
  
        # Iterate over the string
        for j in range(i, len(str), 1):
            flag = 1
  
            # Check for palindrome
            for k in range((j - i + 1) // 2):
                if (str[i + k] != str[j - k]):
                    flag = 0
  
            # If string [i, j - i + 1]
            # is palindromic
            if (flag != 0 and
               (j - i + 1) > maxLength):
                start = i
                maxLength = j - i + 1
             
    # Return length of LPS
    return maxLength
 
# Driver Code
 
# Given string
str = "forgeeksskeegfor"
  
# Function call
print(longestPalSubstr(str))
 
# This code is contributed by code_hunt

C#

// C# program for the above approach 
using System;
 
class GFG{
  
// Function to obtain the length of
// the longest palindromic substring
static int longestPalSubstr(string str)
{
     
    // Length of given string
    int n = str.Length;
  
    // Stores the maximum length
    int maxLength = 1, start = 0;
  
    // Iterate over the string
    for(int i = 0; i < str.Length; i++)
    {
         
        // Iterate over the string
        for(int j = i; j < str.Length; j++)
        {
            int flag = 1;
  
            // Check for palindrome
            for(int k = 0;
                    k < (j - i + 1) / 2; k++)
                if (str[i + k] != str[j - k])
                    flag = 0;
  
            // If string [i, j - i + 1]
            // is palindromic
            if (flag != 0 &&
               (j - i + 1) > maxLength)
            {
                start = i;
                maxLength = j - i + 1;
            }
        }
    }
  
    // Return length of LPS
    return maxLength;
}
  
// Driver Code
public static void Main ()
{
     
    // Given string
    string str = "forgeeksskeegfor";
  
    // Function call
    Console.Write(longestPalSubstr(str));
}
}
 
// This code is contributed by code_hunt

输出：

10

时间复杂度： O(N ³ )，其中 N 是给定字符串的长度。
辅助空间： O(N)

动态规划方法：可以通过存储重叠子问题的结果来优化上述方法。这个想法类似于这篇文章。以下是步骤：

1. 维护一个自底向上填充的布尔表[N][N] 。
2. 如果子串是回文，则table[i][j] 的值为真，否则为假。
3. 要计算table[i][j] ，请检查table[i + 1][j – 1]的值，如果该值为真且str[i]与str[j]相同，则更新table[i] [j]真的。
4. 否则， table[i][j] 的值将更新为 false。

下面是字符串“geeks”的说明：

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 length of
// the longest palindromic substring
int longestPalSubstr(string str)
{
    // Length of string str
    int n = str.size();
 
    // Stores the dp states
    bool table[n][n];
 
    // Initialise table[][] as false
    memset(table, 0, sizeof(table));
 
    // All substrings of length 1
    // are palindromes
    int maxLength = 1;
 
    for (int i = 0; i < n; ++i)
        table[i][i] = true;
 
    // Check for sub-string of length 2
    int start = 0;
 
    for (int i = 0; i < n - 1; ++i) {
 
        // If adjacent character are same
        if (str[i] == str[i + 1]) {
 
            // Update table[i][i + 1]
            table[i][i + 1] = true;
            start = i;
            maxLength = 2;
        }
    }
 
    // Check for lengths greater than 2
    // k is length of substring
    for (int k = 3; k <= n; ++k) {
 
        // Fix the starting index
        for (int i = 0; i < n - k + 1; ++i) {
 
            // Ending index of substring
            // of length k
            int j = i + k - 1;
 
            // Check for palindromic
            // substring str[i, j]
            if (table[i + 1][j - 1]
                && str[i] == str[j]) {
 
                // Mark true
                table[i][j] = true;
 
                // Update the maximum length
                if (k > maxLength) {
                    start = i;
                    maxLength = k;
                }
            }
        }
    }
 
    // Return length of LPS
    return maxLength;
}
 
// Driver Code
int main()
{
    // Given string str
    string str = "forgeeksskeegfor";
 
    // Function Call
    cout << longestPalSubstr(str);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the length of
// the longest palindromic subString
static int longestPalSubstr(String str)
{
     
    // Length of String str
    int n = str.length();
 
    // Stores the dp states
    boolean [][]table = new boolean[n][n];
 
    // All subStrings of length 1
    // are palindromes
    int maxLength = 1;
 
    for(int i = 0; i < n; ++i)
        table[i][i] = true;
 
    // Check for sub-String of length 2
    int start = 0;
 
    for(int i = 0; i < n - 1; ++i)
    {
         
        // If adjacent character are same
        if (str.charAt(i) == str.charAt(i + 1))
        {
             
            // Update table[i][i + 1]
            table[i][i + 1] = true;
            start = i;
            maxLength = 2;
        }
    }
 
    // Check for lengths greater than 2
    // k is length of subString
    for(int k = 3; k <= n; ++k)
    {
         
        // Fix the starting index
        for(int i = 0; i < n - k + 1; ++i)
        {
             
            // Ending index of subString
            // of length k
            int j = i + k - 1;
 
            // Check for palindromic
            // subString str[i, j]
            if (table[i + 1][j - 1] &&
                str.charAt(i) == str.charAt(j))
            {
                 
                // Mark true
                table[i][j] = true;
 
                // Update the maximum length
                if (k > maxLength)
                {
                    start = i;
                    maxLength = k;
                }
            }
        }
    }
     
    // Return length of LPS
    return maxLength;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given String str
    String str = "forgeeksskeegfor";
 
    // Function Call
    System.out.print(longestPalSubstr(str));
}
}
 
// This code is contributed by Amit Katiyar

C＃

// C# program for
// the above approach
using System;
class GFG{
 
// Function to find the length of
// the longest palindromic subString
static int longestPalSubstr(String str)
{
  // Length of String str
  int n = str.Length;
 
  // Stores the dp states
  bool [,]table = new bool[n, n];
 
  // All subStrings of length 1
  // are palindromes
  int maxLength = 1;
 
  for(int i = 0; i < n; ++i)
    table[i, i] = true;
 
  // Check for sub-String
  // of length 2
  int start = 0;
 
  for(int i = 0; i < n - 1; ++i)
  {
    // If adjacent character are same
    if (str[i] == str[i + 1])
    {
      // Update table[i,i + 1]
      table[i, i + 1] = true;
      start = i;
      maxLength = 2;
    }
  }
 
  // Check for lengths greater than 2
  // k is length of subString
  for(int k = 3; k <= n; ++k)
  {
    // Fix the starting index
    for(int i = 0; i < n - k + 1; ++i)
    {
      // Ending index of subString
      // of length k
      int j = i + k - 1;
 
      // Check for palindromic
      // subString str[i, j]
      if (table[i + 1, j - 1] &&
          str[i] == str[j])
      {
        // Mark true
        table[i, j] = true;
 
        // Update the maximum length
        if (k > maxLength)
        {
          start = i;
          maxLength = k;
        }
      }
    }
  }
 
  // Return length of LPS
  return maxLength;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given String str
  String str = "forgeeksskeegfor";
 
  // Function Call
  Console.Write(longestPalSubstr(str));
}
}
 
// This code is contributed by Rajput-Ji

蟒蛇3

# Python program for the above approach
 
 
# Function to find the length of
# the longest palindromic subString
def longestPalSubstr(str):
    # Length of String str
    n = len(str);
 
    # Stores the dp states
    table = [[False for i in range(n)] for j in range(n)];
 
    # All subStrings of length 1
    # are palindromes
    maxLength = 1;
 
    for i in range(n):
        table[i][i] = True;
 
    # Check for sub-String of length 2
    start = 0;
 
    for i in range(n - 1):
 
        # If adjacent character are same
        if (str[i] == str[i + 1]):
            # Update table[i][i + 1]
            table[i][i + 1] = True;
            start = i;
            maxLength = 2;
 
    # Check for lengths greater than 2
    # k is length of subString
    for k in range(3, n + 1):
 
        # Fix the starting index
        for i in range(n - k + 1):
 
            # Ending index of subString
            # of length k
            j = i + k - 1;
 
            # Check for palindromic
            # subString str[i, j]
            if (table[i + 1][j - 1] and str[i] == str[j]):
 
                # Mark True
                table[i][j] = True;
 
                # Update the maximum length
                if (k > maxLength):
                    start = i;
                    maxLength = k;
 
    # Return length of LPS
    return maxLength;
 
 
# Driver Code
if __name__ == '__main__':
    # Given String str
    str = "forgeeksskeegfor";
 
    # Function Call
    print(longestPalSubstr(str));
 
# This code is contributed by 29AjayKumar

输出：

10

时间复杂度： O(N ² )，其中 N 是给定字符串的长度。
辅助空间： O(N)

高效的方法：为了优化上述方法，想法是使用 Manacher 算法。通过使用该算法，对于每个字符c ，可以找到以c为中心的最长回文子串，其长度为奇数。但是最长的回文子串也可以具有没有任何中心的偶数长度。因此，一些特殊字符的每个字符之间增加。

For example, if the given string is “abababc” then it will become “$#a#b#a#b#a#b#c#@”. Now, notice that in this case, for each character c, the longest palindromic substring with the center c will have an odd length.

以下是步骤：

1. 如上所述在给定的字符串S 中添加特殊字符，并使其长度为N 。
2. 用0初始化数组d[] 、 center 和r ，其中d[i]存储回文左侧部分的长度，其中S[i]是中心， r表示最右边的访问边界，center 表示当前索引字符是最右边边界的中心。
3. 在遍历字符串S 时，对于每个索引i ，如果i 小于 r则它的答案之前已经计算过，并且d[i]可以设置为等于以i为中心的字符镜像的答案，可以计算为(2*center – i) 。
4. 现在，检查 r 之后是否有一些字符使回文变得更长。
5. 如果(i + d[i]) 大于 r ，则更新 r = (i + d[i])并以i为中心。
6. 找到以每个字符c为中心的最长回文后，打印(2*d[i] + 1)/2的最大值，其中0 ≤ i < N，因为d[i]只存储了回文的左边部分。

下面是上述方法的实现：

蟒蛇3

# Python program for the above approach
 
# Function that placed '#' intermediately
# before and after each character
def UpdatedString(string):
 
    newString = ['#']
 
# Traverse the string
    for char in string:
        newString += [char, '#']
 
# Return the string
    return ''.join(newString)
 
# Function that finds the length of
# the longest palindromic substring
def Manacher(string):
 
    # Update the string
    string = UpdatedString(string)
 
    # Stores the longest proper prefix
    # which is also a suffix
    LPS = [0 for _ in range(len(string))]
    C = 0
    R = 0
 
    for i in range(len(string)):
        imir = 2 * C - i
 
        # Find the minimum length of
        # the palindrome
        if R > i:
            LPS[i] = min(R-i, LPS[imir])
        else:
 
            # Find the actual length of
            # the palindrome
            LPS[i] = 0
 
        # Exception Handling
        try:
            while string[i + 1 + LPS[i]] \
            == string[i - 1 - LPS[i]]:
                LPS[i] += 1
        except:
            pass
 
        # Update C and R
        if i + LPS[i] > R:
            C = i
            R = i + LPS[i]
 
    r, c = max(LPS), LPS.index(max(LPS))
 
    # Return the length r
    return r
 
 
# Driver code
 
# Given string str
str = "forgeeksskeegfor"
 
# Function Call
print(Manacher(str))

输出：

10

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)

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