通过连接等长前缀和子串可以获得的回文数

先决条件： Z 算法

给定一个字符串S ，任务是找到在执行给定步骤后可以形成的最大回文数：

1. 选择一个非空前缀P和一个等长的非空子串T。
2. 反转P或T并将它们连接起来。

注意： P 和 T 可以重叠。

例子：

Input: S = “abab”
Output: 6
Explanation: 
Consider prefix as S[0 : 1] (= “ab“) and reverse the substring S[2 : 3]. Resultant string after concatenating prefix with reversed substring is “abba”, which is a palindrome.
All possible prefixes are [“a”, “ab”, ”aba”, ”abab”]. 
All possible suffixes are [“a”, ”b”, ”a”, ”b”, ”ab”, ”ba”, ”ab”, ”aba”, ”bab”, ”abab”].
Therefore, all possible palindromes are [“aa”, ”aa”, ”abba”, ”abba”, ”abaaba”, ”ababbaba”].

Input: S = “abcd”
Output: 4

朴素的方法：生成所有可能的前缀和所有可能的子字符串。连接所有等长前缀和子串(反转后)对，并检查获得的字符串是否为回文。

时间复杂度： O(N ³ )
辅助空间： O(N ² )

高效的方法：可以根据观察到的连接字符串的长度总是偶数来优化上述方法。因此，只需要考虑前半部分是后半部分镜像的那些字符串。因此，问题简化为为每个可能的前缀计算与该前缀相等的子串，这可以使用 Z-function 来完成。

The z-function over a string will return an array Z[] where Z[i] represents the length of the longest prefix that same as the sub-string starting at position i and ending at position i+Z[i]. Hence, the count will be $\sum Z[i]+1$.

请按照以下步骤解决问题：

1. 这个想法是维护一个区间[l, r] ，它是具有最大r的区间，使得[l, r]是一个前缀子串(子串也是前缀)。
2. 如果i ≤ r ，则Z[i]等于r – i + 1和Z[i – 1] 的最小值。
3. 现在，增量Z [I]，直到我+ Z [i]是小于N和字符在指数Z [i]和I + Z [I]是给定字符串中相等。
4. 现在，如果i + Z[i] – 1 > r ，则将l设置为i ，将r 设置为i + Z[i] – 1 。
5. 对i从0到N-1重复上述步骤。
6. 答案是每个i (0 ≤ i ≤ N-1)的Z[i] + 1的总和。

下面是上述方法的实现：

C++14

// C++ program the above approach
#include 
using namespace std;
 
// Function to calculate the
// number of palindromes
int countPalindromes(string S)
{
    int N = (int)S.length();
    vector Z(N);
 
    // Calculation of Z-array
    int l = 0, r = 0;
    for (int i = 1; i < N; i++) {
 
        if (i <= r)
            Z[i] = min(r - i + 1, Z[i - l]);
 
        while (i + Z[i] < N
               && S[Z[i]] == S[i + Z[i]]) {
            Z[i]++;
        }
 
        if (i + Z[i] - 1 > r) {
 
            l = i;
            r = i + Z[i] - 1;
        }
    }
 
    // Calculation of sigma(Z[i]+1)
    int sum = 0;
    for (int i = 0; i < Z.size(); i++) {
        sum += Z[i] + 1;
    }
 
    // Return the count
    return sum;
}
 
// Driver Code
int main()
{
    // Given String
    string S = "abab";
    cout << countPalindromes(S);
 
    return 0;
}

Java

// Java program for the above approach 
import java.util.*;
 
class GFG{
     
// Function to calculate the
// number of palindromes
static int countPalindromes(String S)
{
    int N = (int)S.length();
    int[] Z = new int[(N)];
     
    // Calculation of Z-array
    int l = 0, r = 0;
     
    for(int i = 1; i < N; i++)
    {
        if (i <= r)
            Z[i] = Math.min(r - i + 1,
                              Z[i - l]);
 
        while (i + Z[i] < N &&
               S.charAt(Z[i]) ==
               S.charAt(i + Z[i]))
        {
            Z[i]++;
        }
 
        if (i + Z[i] - 1 > r)
        {
            l = i;
            r = i + Z[i] - 1;
        }
    }
 
    // Calculation of sigma(Z[i]+1)
    int sum = 0;
     
    for(int i = 0; i < Z.length; i++)
    {
        sum += Z[i] + 1;
    }
     
    // Return the count
    return sum;
}
 
// Driver Code   
public static void main (String[] args)   
{ 
     
    // Given String
    String S = "abab";
     
    System.out.println(countPalindromes(S));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program the above approach
 
# Function to calculate the
# number of palindromes
def countPalindrome(S):
    N = len(S)
    Z = [0] * N
     
    # Calculation of Z-array
    l = 0
    r = 0
    for i in range(1, N):
        if i <= r:
            Z[i] = min(r - i + 1, Z[i - 1])
        while((i + Z[i]) < N and (S[Z[i]] == S[i + Z[i]])):
            Z[i] += 1
        if ((i + Z[i] - 1) > r):
            l = ir = i + Z[i] - 1
             
    # Calculation of sigma(Z[i]+1)
    sum = 0
    for i in range(0, len(Z)):
        sum += Z[i] + 1
     
    # return the count
    return sum
 
# Driver code
 
# Given String
S = "abab"
print(countPalindrome(S))
 
# This code is contributed by virusbuddah

C#

// C# program for the above approach 
using System;
 
public class GFG{
     
// Function to calculate the
// number of palindromes
static int countPalindromes(String S)
{
    int N = (int)S.Length;
    int[] Z = new int[(N)];
     
    // Calculation of Z-array
    int l = 0, r = 0;
     
    for(int i = 1; i < N; i++)
    {
        if (i <= r)
            Z[i] = Math.Min(r - i + 1,
                              Z[i - l]);
        while (i + Z[i] < N &&
               S[Z[i]] ==
               S[i + Z[i]])
        {
            Z[i]++;
        }
 
        if (i + Z[i] - 1 > r)
        {
            l = i;
            r = i + Z[i] - 1;
        }
    }
 
    // Calculation of sigma(Z[i]+1)
    int sum = 0;
     
    for(int i = 0; i < Z.Length; i++)
    {
        sum += Z[i] + 1;
    }
     
    // Return the count
    return sum;
}
 
// Driver Code   
public static void Main(String[] args)   
{ 
     
    // Given String
    String S = "abab";
     
    Console.WriteLine(countPalindromes(S));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascript program the above approach
 
// Function to calculate the
// number of palindromes
function countPalindromes(S)
{
    var N = S.length;
    var Z = Array(N).fill(0);
 
    // Calculation of Z-array
    var l = 0, r = 0;
    for (var i = 1; i < N; i++) {
 
        if (i <= r)
            Z[i] = Math.min(r - i + 1, Z[i - l]);
 
        while (i + Z[i] < N
            && S[Z[i]] == S[i + Z[i]]) {
            Z[i]++;
        }
 
        if (i + Z[i] - 1 > r) {
 
            l = i;
            r = i + Z[i] - 1;
        }
    }
 
    // Calculation of sigma(Z[i]+1)
    var sum = 0;
    for (var i = 0; i < Z.length; i++) {
        sum += Z[i] + 1;
    }
 
    // Return the count
    return sum;
}
 
// Driver Code
// Given String
var S = "abab";
document.write( countPalindromes(S));
 

输出：

6

时间复杂度： O(N)
辅助空间： O(N)

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