以X开头和Y结束的最长子字符串

给定一个字符串str ，两个字符X和Y。任务是找到以X开头和Y结尾的最长子字符串的长度。假定始终存在一个以X开头和Y结尾的子字符串。
例子：

Input: str = “QWERTYASDFZXCV”, X = ‘A’, Y = ‘Z’ 
Output: 5 
Explanation: 
The largest substring which start with ‘A’ and end with ‘Z’ = “ASDFZ”. 
Size of the substring = 5.
Input: str = “ZABCZ”, X = ‘Z’, Y = ‘Z’ 
Output: 3 
Explanation: 
The largest substring which start with ‘Z’ and end with ‘Z’ = “ZABCZ”. 
Size of the substring = 5. 
 

天真的方法：天真的方法是从给定字符串的所有子字符串中找到所有以X开头并以Y结束的最大子字符串。
时间复杂度： O(N ² )
辅助空间： O(1)
高效方法：为优化上述方法， X和Y之间的字符数应为最大。因此，使用指针start和end遍历字符串，从起始索引中查找X的第一次出现，从末尾查找Y的最后一次出现。步骤如下：

1. 初始化start = 0和end =字符串的长度– 1 。
2. 从头开始遍历字符串，并找到字符X的第一个匹配项。使其位于索引xPos处。
3. 从头开始遍历字符串，并找到字符Y的最后一次出现。使其位于索引yPos 。
4. 最长子字符串的长度由(yPos – xPos + 1)给出。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function returns length of longest
// substring starting with X and
// ending with Y
int longestSubstring(string str,
                    char X, char Y)
{
    // Length of string
    int N = str.length();
    int start = 0;
    int end = N - 1;
    int xPos = 0;
    int yPos = 0;
 
    // Find the length of the string
    // starting with X from the beginning
    while (true) {
 
        if (str[start] == X) {
            xPos = start;
            break;
        }
        start++;
    }
 
    // Find the length of the string
    // ending with Y from the end
    while (true) {
 
        if (str[end] == Y) {
            yPos = end;
            break;
        }
        end--;
    }
 
    // Longest substring
    int length = (yPos - xPos) + 1;
 
    // Print the length
    cout << length;
}
 
// Driver Code
int main()
{
    // Given string str
    string str = "HASFJGHOGAKZZFEGA";
 
    // Starting and Ending characters
    char X = 'A', Y = 'Z';
 
    // Function Call
    longestSubstring(str, X, Y);
    return 0;
}

Java

// Java program for the above approach
class GFG{
     
// Function returns length of longest
// substring starting with X and
// ending with Y
public static void longestSubstring(String str,
                                    char X, char Y)
{
     
    // Length of string
    int N = str.length();
    int start = 0;
    int end = N - 1;
    int xPos = 0;
    int yPos = 0;
     
    // Find the length of the string
    // starting with X from the beginning
    while (true)
    {
        if (str.charAt(start) == X)
        {
            xPos = start;
            break;
        }
        start++;
    }
     
    // Find the length of the string
    // ending with Y from the end
    while (true)
    {
        if (str.charAt(end) == Y)
        {
            yPos = end;
            break;
        }
        end--;
    }
     
    // Longest substring
    int length = (yPos - xPos) + 1;
     
    // Print the length
    System.out.print(length);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given string str
    String str = "HASFJGHOGAKZZFEGA";
     
    // Starting and Ending characters
    char X = 'A', Y = 'Z';
     
    // Function Call
    longestSubstring(str, X, Y);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
 
# Function returns length of longest
# substring starting with X and
# ending with Y
def longestSubstring(str, X, Y):
     
    # Length of string
    N = len(str)
     
    start = 0
    end = N - 1
    xPos = 0
    yPos = 0
 
    # Find the length of the string
    # starting with X from the beginning
    while (True):
        if (str[start] == X):
            xPos = start
            break
             
        start += 1
 
    # Find the length of the string
    # ending with Y from the end
    while (True):
        if (str[end] == Y):
            yPos = end
            break
         
        end -= 1
 
    # Longest substring
    length = (yPos - xPos) + 1
 
    # Print the length
    print(length)
 
# Driver Code
if __name__ == "__main__":
     
    # Given string str
    str = "HASFJGHOGAKZZFEGA"
 
    # Starting and Ending characters
    X = 'A'
    Y = 'Z'
 
    # Function Call
    longestSubstring(str, X, Y)
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach 
using System;
 
class GFG{
     
// Function returns length of longest
// substring starting with X and
// ending with Y
static void longestSubstring(string str,
                             char X, char Y)
{
     
    // Length of string
    int N = str.Length;
    int start = 0;
    int end = N - 1;
    int xPos = 0;
    int yPos = 0;
     
    // Find the length of the string
    // starting with X from the beginning
    while (true)
    {
        if (str[start] == X)
        {
            xPos = start;
            break;
        }
        start++;
    }
     
    // Find the length of the string
    // ending with Y from the end
    while (true)
    {
        if (str[end] == Y)
        {
            yPos = end;
            break;
        }
        end--;
    }
     
    // Longest substring
    int length = (yPos - xPos) + 1;
     
    // Print the length
    Console.Write(length);
}
 
// Driver code
public static void Main()
{
     
    // Given string str
    string str = "HASFJGHOGAKZZFEGA";
     
    // Starting and Ending characters
    char X = 'A', Y = 'Z';
     
    // Function call
    longestSubstring(str, X, Y);
}
}
 
// This code is contributed by sanjoy_62

Javascript

      // JavaScript program for the above approach
      // Function returns length of longest
      // substring starting with X and
      // ending with Y
      function longestSubstring(str, X, Y) {
        // Length of string
        var N = str.length;
        var start = 0;
        var end = N - 1;
        var xPos = 0;
        var yPos = 0;
 
        // Find the length of the string
        // starting with X from the beginning
        while (true) {
          if (str[start] === X) {
            xPos = start;
            break;
          }
          start++;
        }
 
        // Find the length of the string
        // ending with Y from the end
        while (true) {
          if (str[end] === Y) {
            yPos = end;
            break;
          }
          end--;
        }
 
        // Longest substring
        var length = yPos - xPos + 1;
 
        // Print the length
        document.write(length);
      }
 
      // Driver code
      // Given string str
      var str = "HASFJGHOGAKZZFEGA";
 
      // Starting and Ending characters
      var X = "A",
        Y = "Z";
      // Function call
      longestSubstring(str, X, Y);
    

输出：

12

时间复杂度： O(N)
辅助空间： O(1)

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