Understanding how to search for string within another string

Question

I'm trying to write for code that searches for the first occurrence of a string (a needle) within another string (a haystack) and return the index of the first character in the matching string.

I've found the following solution online, but I'm having trouble understanding how it works. Can someone please explain to me how it functions?

private static int strStr(String needle, String haystack)
{
    for (int i=0;;i++) {  //
        for (int j=0;;j++) {
            if (j== needle.length()) return i;
            if (i + j == haystack.length()) return -1;
            System.out.println("needle.charAt(j) "+j+" "+needle.charAt(j));
            System.out.println("haystack.charAt(i+j) "+(i+j)+" "+haystack.charAt(i+j));
            if (needle.charAt(j) != haystack.charAt(i + j)) break;
        }
    }
}

Answer 1

private static int strStr(String needle, String haystack)
{
    for (int i = 0 ; ; i++)
    {
        for (int j = 0 ; ; j++)
        {
            if (j == needle.length())
            {
                return i;
            }

            if (i + j == haystack.length())
            {
                return -1;
            }

            System.out.println("needle.charAt(j) " + j + " " + needle.charAt(j));
            System.out.println("haystack.charAt(i+j) " + (i + j) + " " + haystack.charAt(i+j));

            if (needle.charAt(j) != haystack.charAt(i + j))
            {
                break;
            }
        }
    }
}

First, let's establish a few things:

• Java starts indices at 0, so needle.charAt(0) is the first character in the string. needle.charAt(3) is the fourth character in the string.
• The line for(int i = 0 ; ; i++) increments i by one through each iteration, and the for-loop will not cause the loop to stop.

It's easiest to tackle this with an example. Let's use a needle named "hip" and a haystack named "chips". Since Java indices start at 0, we would expect that the method will return 1 (representing the second character), since that's the first character of "hip" within "chips".

• When we enter the method, we go through the first for-loop. i = 0
• The next line takes us into another for-loop. i = 0, j = 0
• j = 0 does not equal needle.length() = 3
• i + j = 0 + 0 = 0 does not equal haystack.length() = 5
• needle.charAt(j = 0) is "h", haystack.charAt(i + j = 0 + 0 = 0) is "c". Since "h" is not equal to "c", we break out of the j-for-loop. The break keyword only breaks out of the current loop, so while we stopped executing the j for-loop, we're still within the i-for-loop.
• We start the next iteration. i = 1
• The j for-loop restarts at 0. i = 1, j = 0
• j = 0 does not equal needle.length() = 3
• i + j = 1 + 0 = 1 does not equal haystack.length() = 5
• needle.charAt(j = 0) is "h", haystack.charAt(i + j = 1 + 0 = 1) is "h". Since these are equal, we don't break out of the j-for-loop.
• We start the next iteration of the j for-loop. i = 1, j = 1
• j = 1 does not equal needle.length() = 3
• i + j = 1 + 1 = 2 does not equal haystack.length() = 5
• needle.charAt(j = 1) is "i", haystack.charAt(i + j = 1 + 1 = 2) is "i". Since these are equal, we don't break out of the j-for-loop.
• We start the next iteration of the j for-loop. i = 1, j = 2
• j = 2 does not equal needle.length() = 3
• i + j = 1 + 2 = 3 does not equal haystack.length() = 5
• needle.charAt(j = 2) is "p", haystack.charAt(i + j = 1 + 2) is "p". Since these are equal, we don't break out of the j-for-loop.
• We start the next iteration of the j for-loop. i = 1, j = 3
• j = 3 does equal needle.length() = 3, so we return i = 1.

As we expected, we got 1 back from this function, since "hip" is contained within "chips" starting at position 1 (zero-indexed).

That's all well and good, but what about that i + j == haystack.length() line?

Let's use "ben" as the needle and "bear" as the haystack (we should get back -1, since "ben" does not appear in the word "bear").

         /- needle.length()
         |   /- haystack.length()
         |   |   /- needle.charAt(j)
         |   |   |   /- haystack.charAt(i + j)
         |   |   |   |
 i | j |n.l|h.l|n.c|h.c| result
---+---+---+---+---+---+------------------
 0 | 0 | 3 | 4 | b | b | continue
 0 | 1 | 3 | 4 | e | e | continue
 0 | 2 | 3 | 4 | n | a | break j
 1 | 0 | 3 | 4 | b | e | break j
 2 | 0 | 3 | 4 | b | a | break j
 3 | 0 | 3 | 4 | b | r | break j
 4 | 0 | 3 | 4 |   |   | i + j == haystack.length(), return -1

Answer 2

you can use contains() or indexOf() for that. for example:

String str1 = "This is string of words";
String str2 = "string";
int pos = str1.indexOf(str2);

in pos will int with position. Also you can add toUpperCase(), in this case search will be without letter register.