Learn how to efficiently find a substring using recursion in Java, focusing on the essential String methods.
---
This video is based on the question https://stackoverflow.com/q/67904001/ asked by the user 'Pompa' ( https://stackoverflow.com/u/16115274/ ) and on the answer https://stackoverflow.com/a/67905357/ provided by the user 'Stef' ( https://stackoverflow.com/u/3080723/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.
Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: Finding a substring of a string with recursion using only the next Methods of class String : charAt(),substring(),length()
Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l...
The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license.
If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com.
---
How to Find a Substring in Java Using Recursion: A Step-by-Step Guide
When working with strings in Java, you may encounter situations where you need to determine if one string (let's call it b) is a substring of another string (a). Traditionally, this can be achieved through built-in methods, but what if you're looking to implement it recursively? This guide dives into how to accomplish this specifically using the charAt(), substring(), and length() methods from the String class.
The Problem: Checking for a Substring
Let's say you have the following pairs of strings:
Example 1: a = "welcome", b = "come" - Output: true
Example 2: a = "hi", b = "hi" - Output: false
Example 3: a = "goal", b = "gl" - Output: false
You may have noticed that a common issue with naive substring searches is returning true when the letters of both strings overlap in unfortunate ways. For example, the pair a = "subtring", b = "string" might incorrectly return true when it's not a valid substring search.
The Solution: Recursion with Two Functions
The key to solving this problem effectively lies in modifying our approach by using two recursive functions: one to check for a generic substring and another specialized function to check for prefixes (substrings that start at position 0 of the string). Here’s how we can structure it.
Step 1: Verify Input Strings
The very first thing you should check is the length ratio between the two strings:
If b's length is greater than or equal to a's length, it cannot be a substring, thus returning false.
If b is empty, it is trivially a substring, leading to a return of true.
Step 2: Implementing the isPrefix Function
This function is simpler—it checks if b is located at the start of a:
[[See Video to Reveal this Text or Code Snippet]]
This approach ensures that you only check for matches directly at the beginning of a.
Step 3: Implementing the isSubstring Function
Next, we build upon our prefix check using the isSubstring function, which searches b throughout all of a:
[[See Video to Reveal this Text or Code Snippet]]
Combining the Functions
With both functions defined, your full code to find a substring recursively will look like this:
[[See Video to Reveal this Text or Code Snippet]]
Conclusion
In conclusion, finding a substring recursively can be streamlined by separating concerns into prefix checking and substring searching. By using the two functions isPrefix and isSubstring, you can effectively circumvent false positives while maintaining a clear and concise recursive structure. Now, you can implement this technique in your Java programming projects!
Информация по комментариям в разработке