Simple String matching in java

Question

I am trying to implement simple String matching. The algorithm should return 1 if the source String contains the pattern String. I cannot understand why it is returning -1 on the following inputs

    String source = "aababba";
    String pattern = "abba";

Here is my implementation:

public static int findMatch(String source, String pattern)
{
    int j = 0,  pos = -1;
    boolean matched = false;

    if(source.length() < pattern.length())
        return -1;


    for(int i = 0; i < (source.length() - pattern.length()); i++)
    {
        if(source.charAt(i) == pattern.charAt(j))
            j++;
        else
            j = 0;
        if(j == pattern.length())
        {
            matched = true;
            break;
        }           
    }

    if(matched)
        return 1;       
    return -1;  
}   

EDIT: As many of you suggested, the culprit was the for loop. I should have made it as follows. The rest of the code is the same. Other solutions are also possible as shown in the answers.

for(int i = 0; i <= (source.length() - pattern.length()); i++)
    {
        if(source.charAt(i+j) == pattern.charAt(j))
        {

Answer 1

You are only checking three characters

source.length() - pattern.length()

So j will never be equal to pattern.length . You need to check the entire source .

for(int i = 0; i < source.length(); i++)
{
  if(source.charAt(i) == pattern.charAt(j))
  ...
}

---

Regarding the Solution

for(int i = 0; i <= (source.length() - pattern.length()); i++)
    {
        if(source.charAt(i+j) == pattern.charAt(j))
        {

Why have an additional subtraction and addition plus more complicated code when you can just step through the length of source?

Answer 2

The source of an issue is that you've limited the upper bound of i variable by (source.length() - pattern.length()) . The for loop you've written can't check all the characters of source string, thus it'll return -1 even for some pairs of strings where it's possible to find a match.
Solution: rewrite the for loop like so:

for(int i = 0; i < source.length(); i++)

Answer 3

源的长度为7。模式的长度为4。您的语句i < (source.length() - pattern.length()对于i> 2而言将不成立，因此循环不会足够“远”地运行。

Answer 4

您可以轻松地使用String.contains方法检查字符串是否包含模式。

Answer 5

I think you already know the reason why it return wrong. You can try this:

j = 0；
while（j <= (source.length() - pattern.length())）{

    for (i = pattern.length(); i >= 0 && pattern[i] == source[i + j]; --i);

    if (i < 0)) {
        matched = true;
        break;
    }
    else j++；
}

about string matching, there are many methods, you can use KMP, BM and Sunday and so on, and performance comparation is: KMP < BM < Sunday, you can try these and they are very userful.

Answer 6

By using this statement (source.length() - pattern.length()) you are looping on the string only on the three first chars of the source. So I will never find the entire pattern.

Try using pattern.length() instead, or better try compute the minimal value of both pattern.length() and source.length()

Answer 7

I'll assume it should return 1 because "aababba" contains "abba" at the end.

This, from your edit (solution) will return 1 :

for(int i = 0; i <= (source.length() - pattern.length()); i++) {
    if(source.charAt(i+j) == pattern.charAt(j)) {

But from my assumption, you should be checking through every letter in source so I would use:

for(int i = 0; i < source.length(); i++) {
    if(source.charAt(i) == pattern.charAt(j)) {

Because why ignore the last bunch of characters?

But maybe I'm just assuming wrong what you want your algorithm to do because from what I've assumed, you could just use:

if (source.contains(pattern))
    return 1;
else
    return -1;