任何一对相似字符之间的最长子串

给定一个字符串S ，任务是找到任何一对相同字符之间的最长子字符串的长度。

例子：

Input: S = “accabbacc” 
Output: 6 
Explanation: The longest substring satisfying the required conditions is “cabbac”, which lies between S[1](= ‘c’) and s[8](= ‘c’).

Input: S = “aab” 
Output: 0

处理方法：按照以下步骤解决问题：

1. 初始化两个变量res和diff分别存储相同字符对之间最长子串的长度和当前子串的长度。
2. 从左到右迭代字符串的字符。
3. 遍历右侧剩余的字符串，从右到左直到当前字符。
4. 如果得到两个相等的字符，即S[i] = S[j], ，将它们之间的子串的长度存储在diff 中。
5. 将res的值更新为max(res, diff)。以便res存储迄今为止获得的所需类型的最长子字符串。
6. 完成字符串遍历后，打印res作为所需答案。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find the longest substring
// between pair of repetitions of the same character
int longestbetweenequalcharacters(string S)
{
 
    // Length of the string
    int n = S.length();
 
    // Stores the maximum length and
    // length of current substring
    // satisfying given conditions
    int res = -1, diff = -1;
 
    // Traverse the string
    for (int i = 0; i < n - 1; i++) {
 
        // Search for repetition of S[i]
        for (int j = n - 1; j > i; j--) {
 
            // If repetition occurs
            if (S[i] == S[j]) {
 
                // Store the length of
                // the substring
                diff = j - i - 1;
 
                // Update maximum length of
                // substring obtained
                res = max(diff, res);
            }
        }
    }
 
    // Return obtained maximum length
    return res;
}
 
// Driver Code
int main()
{
    string s = "accabbacc";
    cout << longestbetweenequalcharacters(s);
}

Java

// Java program to implement
// the above approach
class GFG{
      
// Function to find the longest substring
// between pair of repetitions of the
// same character
static int longestbetweenequalcharacters(String S)
{
     
    // Length of the string
    int n = S.length();
  
    // Stores the maximum length and
    // length of current substring
    // satisfying given conditions
    int res = -1, diff = -1;
  
    // Traverse the string
    for(int i = 0; i < n - 1; i++)
    {
         
        // Search for repetition of S[i]
        for(int j = n - 1; j > i; j--)
        {
             
            // If repetition occurs
            if (S.charAt(i) == S.charAt(j))
            {
                 
                // Store the length of
                // the substring
                diff = j - i - 1;
  
                // Update maximum length of
                // substring obtained
                res = Math.max(diff, res);
            }
        }
    }
  
    // Return obtained maximum length
    return res;
}
  
// Driver code
public static void main(String[] args)
{
     
    String s = "accabbacc";
     
    System.out.println(
        longestbetweenequalcharacters(s));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program to implement
# the above approach
 
# Function to find the longest
# substring between pair of
# repetitions of the same character
def longestbetweenequalcharacters(S):
     
    # Length of the string
    n = len(S)
 
    # Stores the maximum length and
    # length of current substring
    # satisfying given conditions
    res = -1
    diff = -1
 
    # Traverse the string
    for i in range(n - 1):
         
        # Search for repetition of S[i]
        for j in range(n - 1, i, -1):
             
            # If repetition occurs
            if (S[i] == S[j]):
 
                # Store the length of
                # the substring
                diff = j - i - 1
 
                # Update maximum length of
                # substring obtained
                res = max(diff, res)
 
    # Return obtained maximum length
    return res
 
# Driver Code
if __name__ == '__main__':
     
    s = "accabbacc"
     
    print(longestbetweenequalcharacters(s))
 
# This code is contributed by doreamon_

C#

// C# program to implement
// the above approach 
using System;
 
class GFG{
       
// Function to find the longest substring
// between pair of repetitions of the
// same character
static int longestbetweenequalcharacters(String S)
{
     
    // Length of the string
    int n = S.Length;
   
    // Stores the maximum length and
    // length of current substring
    // satisfying given conditions
    int res = -1, diff = -1;
   
    // Traverse the string
    for(int i = 0; i < n - 1; i++)
    {
         
        // Search for repetition of S[i]
        for(int j = n - 1; j > i; j--)
        {
             
            // If repetition occurs
            if (S[i] == S[j])
            {
                 
                // Store the length of
                // the substring
                diff = j - i - 1;
   
                // Update maximum length of
                // substring obtained
                res = Math.Max(diff, res);
            }
        }
    }
   
    // Return obtained maximum length
    return res;
}
   
// Driver code
public static void Main()
{
    string s = "accabbacc";
      
    Console.WriteLine(
        longestbetweenequalcharacters(s));
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
// javascript program to implement
// the above approach
 
      
// Function to find the longest substring
// between pair of repetitions of the
// same character
function longestbetweenequalcharacters(S)
{
     
    // Length of the string
    var n = S.length;
  
    // Stores the maximum length and
    // length of current substring
    // satisfying given conditions
    var res = -1, diff = -1;
  
    // Traverse the string
    for(i = 0; i < n - 1; i++)
    {
         
        // Search for repetition of S[i]
        for(j = n - 1; j > i; j--)
        {
             
            // If repetition occurs
            if (S.charAt(i) == S.charAt(j))
            {
                 
                // Store the length of
                // the substring
                diff = j - i - 1;
  
                // Update maximum length of
                // substring obtained
                res = Math.max(diff, res);
            }
        }
    }
  
    // Return obtained maximum length
    return res;
}
  
// Driver code
 
var s = "accabbacc";
 
document.write(
    longestbetweenequalcharacters(s));
 
// This code contributed by shikhasingrajput
 

输出：

6

时间复杂度： O(N ² )
辅助空间： O(1)

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