字符串任意两个相同字符之间的最大字符数

给定一个字符串，找到字符串中的任意两个字符之间的最大字符数。如果没有字符重复，则打印 -1。

例子：

Input : str = "abba"
Output : 2
The maximum number of characters are
between two occurrences of 'a'.

Input : str = "baaabcddc"
Output : 3
The maximum number of characters are
between two occurrences of 'b'.

Input : str = "abc"
Output : -1

方法 1（简单）：使用两个嵌套循环。外循环从左到右选择字符，内循环找到最远的出现并跟踪最大值。

C++

// Simple C++ program to find maximum number
// of characters between two occurrences of
// same character
#include 
using namespace std;
 
int maximumChars(string& str)
{
    int n = str.length();
    int res = -1;
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if (str[i] == str[j])
                res = max(res, abs(j - i - 1));
    return res;
}
 
// Driver code
int main()
{
    string str = "abba";
    cout << maximumChars(str);
    return 0;
}

Java

// Simple Java program to find maximum
// number of characters between two
// occurrences of same character
class GFG {
     
    static int maximumChars(String str)
    {
        int n = str.length();
        int res = -1;
         
        for (int i = 0; i < n - 1; i++)
            for (int j = i + 1; j < n; j++)
                if (str.charAt(i) == str.charAt(j))
                    res = Math.max(res,
                         Math.abs(j - i - 1));
                          
        return res;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        String str = "abba";
         
        System.out.println(maximumChars(str));
    }
}
 
// This code is contributed by vt_m.

Python3

# Simple Python3 program to find maximum number
# of characters between two occurrences of
# same character
def maximumChars(str):
    n = len(str)
    res = -1
    for i in range(0, n - 1):
        for j in range(i + 1, n):
            if (str[i] == str[j]):
                res = max(res, abs(j - i - 1))
    return res
 
# Driver code
if __name__ == '__main__':
    str = "abba"
    print(maximumChars(str))
 
# This code is contributed by PrinciRaj1992

C#

// Simple C# program to find maximum
// number of characters between two
// occurrences of same character
using System;
 
public class GFG {
     
    static int maximumChars(string str)
    {
        int n = str.Length;
        int res = -1;
         
        for (int i = 0; i < n - 1; i++)
            for (int j = i + 1; j < n; j++)
                if (str[i] == str[j])
                    res = Math.Max(res,
                         Math.Abs(j - i - 1));
                          
        return res;
    }
     
    // Driver code
    static public void Main ()
    {
        string str = "abba";
         
        Console.WriteLine(maximumChars(str));
    }
}
 
// This code is contributed by vt_m.

Javascript

    // Simple Javascript program to find maximum
    // number of characters between two
    // occurrences of same character
     
    function maximumChars(str)
    {
        let n = str.length;
        let res = -1;
           
        for (let i = 0; i < n - 1; i++)
            for (let j = i + 1; j < n; j++)
                if (str[i] == str[j])
                    res = Math.max(res,
                         Math.abs(j - i - 1));
                            
        return res;
    }
     
    // Driver code
    let str = "abba";
    document.write(maximumChars(str));
     
 
// This code is contributed by unknown2108

输出：

2

时间复杂度： O(n^2)

方法 2（高效）：初始化一个长度为 26 的数组“FIRST”，我们必须在其中存储字符串第一次出现的字母表和另一个长度为 26 的数组“LAST”，我们将存储最后一次出现的字母表在字符串。这里，索引 0 对应于字母 a，1 对应于 b 等等。之后，如果它们不在同一位置，我们将取最后一个和第一个数组之间的差异来找到最大差异。

C++

// Efficient C++ program to find maximum number
// of characters between two occurrences of
// same character
#include 
using namespace std;
 
const int MAX_CHAR = 256;
 
int maximumChars(string& str)
{
    int n = str.length();
    int res = -1;
 
    // Initialize all indexes as -1.
    int firstInd[MAX_CHAR];
    for (int i = 0; i < MAX_CHAR; i++)
        firstInd[i] = -1;
 
    for (int i = 0; i < n; i++) {
        int first_ind = firstInd[str[i]];
 
        // If this is first occurrence
        if (first_ind == -1)
            firstInd[str[i]] = i;
 
        // Else find distance from previous
        // occurrence and update result (if
        // required).
        else
            res = max(res, abs(i - first_ind - 1));
    }
    return res;
}
 
// Driver code
int main()
{
    string str = "abba";
    cout << maximumChars(str);
    return 0;
}

Java

// Efficient java program to find maximum
// number of characters between two
// occurrences of same character
import java.io.*;
 
public class GFG {
 
    static int MAX_CHAR = 256;
     
    static int maximumChars(String str)
    {
        int n = str.length();
        int res = -1;
     
        // Initialize all indexes as -1.
        int []firstInd = new int[MAX_CHAR];
         
        for (int i = 0; i < MAX_CHAR; i++)
            firstInd[i] = -1;
     
        for (int i = 0; i < n; i++)
        {
            int first_ind = firstInd[str.charAt(i)];
     
            // If this is first occurrence
            if (first_ind == -1)
                firstInd[str.charAt(i)] = i;
     
            // Else find distance from previous
            // occurrence and update result (if
            // required).
            else
                res = Math.max(res, Math.abs(i
                                  - first_ind - 1));
        }
         
        return res;
    }
     
    // Driver code
     
    static public void main (String[] args)
    {
        String str = "abba";
         
        System.out.println(maximumChars(str));
    }
}
 
// This code is contributed by vt_m.

蟒蛇3

# Efficient Python3 program to find maximum
# number of characters between two occurrences
# of same character
 
MAX_CHAR = 256
 
def maximumChars(str1):
 
    n = len(str1)
    res = -1
 
    # Initialize all indexes as -1.
    firstInd = [-1 for i in range(MAX_CHAR)]
 
    for i in range(n):
        first_ind = firstInd[ord(str1[i])]
 
        # If this is first occurrence
        if (first_ind == -1):
            firstInd[ord(str1[i])] = i
 
        # Else find distance from previous
        # occurrence and update result (if
        # required).
        else:
            res = max(res, abs(i - first_ind - 1))
     
    return res
 
# Driver code
str1 = "abba"
print(maximumChars(str1))
 
# This code is contributed by Mohit kumar 29

C＃

// Efficient C# program to find maximum
// number of characters between two
// occurrences of same character
using System;
 
public class GFG {
 
    static int MAX_CHAR = 256;
     
    static int maximumChars(string str)
    {
        int n = str.Length;
        int res = -1;
     
        // Initialize all indexes as -1.
        int []firstInd = new int[MAX_CHAR];
         
        for (int i = 0; i < MAX_CHAR; i++)
            firstInd[i] = -1;
     
        for (int i = 0; i < n; i++)
        {
            int first_ind = firstInd[str[i]];
     
            // If this is first occurrence
            if (first_ind == -1)
                firstInd[str[i]] = i;
     
            // Else find distance from previous
            // occurrence and update result (if
            // required).
            else
                res = Math.Max(res, Math.Abs(i
                              - first_ind - 1));
        }
         
        return res;
    }
     
    // Driver code
 
    static public void Main ()
    {
        string str = "abba";
         
        Console.WriteLine(maximumChars(str));
    }
}
 
// This code is contributed by vt_m.

Javascript

    // Efficient javascript program to find maximum
    // number of characters between two
    // occurrences of same character
     
    let MAX_CHAR = 256;
     
    function maximumChars(str)
    {
        let n = str.length;
        let res = -1;
       
        // Initialize all indexes as -1.
        let firstInd = new Array(MAX_CHAR);
           
        for (let i = 0; i < MAX_CHAR; i++)
            firstInd[i] = -1;
       
        for (let i = 0; i < n; i++)
        {
           let first_ind = firstInd[str[i].charCodeAt(0)];
       
            // If this is first occurrence
            if (first_ind == -1)
                firstInd[str[i].charCodeAt(0)] = i;
       
            // Else find distance from previous
            // occurrence and update result (if
            // required).
            else
                res = Math.max(res, Math.abs(i
                                  - first_ind - 1));
        }
           
        return res;
    }
     
    // Driver code
    let str = "abba";
    document.write(maximumChars(str));
     
 
// This code is contributed by patel2127

输出：

2

时间复杂度： O(n)

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