如何比较两个不同长度的字符串以找到相同的子字符串

Question

The problem is: Client accounts are filed under a classification system using codes eg MA400. I need a method that will reset the original MA400 to an updated code such as MA400.4. If the new code has 5 characters to which the original is reset then the method returns true. Not the best wording but that is all I have right now.

It hasn't been specified if the characters need to be in the same order, eg.

    String str = "abc123";
    String newStr = "xyz123abc";

I am assuming they need to be in the same order. So the above strings would only have 3 like characters.

    char[]array = str.toCharArray();
    char[]array2 = newStr.toCharArray();

I am thinking now to use a compareTo method on the two arrays, but I am not sure how this would work exactly. Perhaps I could use a for loop to stop comparing after the final element in the shortest string but not entirely sure if I can do much with that.

I feel like I am going about this in the wrong way and there is a less complicated way to check for like characters in a string?

Answer 1

From what I understand something like this will work. Remember this will only count unique characters. Order does not matter

public static boolean matchingChar(final String st1, final String st2) {

        if(st1 == null || st2 == null || st1.length() < 5  || st2.length() < 5) {
            return false;
        } 

        //This is if you wish unique characters to be counted only
        //Otherwise you can use simple int count = 0
        HashSet<Character> found = new HashSet<Character>();

        //found.size() < 5 so the loop break as soon as the condition is met
        for(int i = 0; i < st1.length() && found.size() < 5; i++) {         
            if(st2.indexOf(st1.charAt(i)) != -1) {
                found.add(st1.charAt(i));
            }
        }

        return found.size() >= 5;
    }