从两个N个或更多的重叠字符串中创建一个字符串？

Question

Say I have two Strings,

str1 = "abcdefg"
str2 = "cdefghijkl"

If a user specified an integer amount, say n = 4.

how do I find if the two Strings overlap by 4 or more characters? In this case they overlap "cdefg" by 5 to make "abcdefghijkl"

If n = 6 then they would not overlap because "bcdefg" != "cdefgh"

The issue I am having is if there is more overlap than user specified.

Answer 1

The issue I am having is if there is more overlap than user specified.

That's where the loop comes in: start at the position defined by user's overlap, and keep incrementing it until (1) you detect an overlap or (2) you run out of characters in one of the strings.

Pseudo-code would look like this:

int userOverlap = ... // Get user overlap
int minLength = Math.min(stringOne.length(), stringTwo.length());
for (int overlap = userOverlap  ; overlap <= minLength; overlap++) {
    if (testOverlap(stringOne, stringTwo, overlap)) {
        return true;
    }
}
return false;

private static bool testOverlap(String a, String b, int overlap) {
    ... // This is your method that tests for one specific overlap
}

Answer 2

The first step is to see if they overlap at all. You could use the following routine to check that:

/**
 * Determine if string 2 overlaps string 1 at all
 * @param st1 The first string to check.
 * @param st2 The second string to check
 * @param n The minimum number of characters.
 * @return The character index in string 2 that 
 *         overlaps with something in string 1
 *         a -1 indicates no overlap
 */
static int st1Pos;
static int st2Pos;
static int getOverlap(String st1, String st2, int n)
{
    st1Pos = -1;
    st2Pos = -1;
    if ((n <= 0) || (st1.length() < n) || (st2.length() < n))
    {
        return -1;
    }
    int index;
    for (int i = 0; i < st1.length() - n; i++)
    {
        String sub = st2.substring(i, i + n);
        index = st2.indexOf(sub);
        if (index >= 0)
        {
            st1Pos = index;
            st2Pos = i;
            return i;
        }
    }
    return -1;
}

Once you do this you st1Pos and st2Pos indicate where the overlap starts. You can then getChars to convert the two strings into character arrays and then start at the locations and compare as shown below:

/**
 * Return the number of characters that the two strings overlap 
 * @param str1
 * @param str2
 * @param min - minimum required overlap
 * @return The number they actually overlap for first detected overlap
 */    
static int numOverlap(String str1, String str2, int min)
{
    int pos = getOverlap(str1, str2, min);
    if (pos < 0)
    {
        return -1;
    }
    char a[] = str1.toCharArray();
    char b[] = str2.toCharArray();
    int i = 0;
    while (a[i + st1Pos] == b[i + st2Pos])
    {
        i++;
        if ((i + st1Pos >= a.length) || (i + st2Pos >= b.length))
        {
            break;
        }
    }
    return i;
}

A few things to note about this implementation. First, I did everything static for ease of explanation - you probably want to make this a class and put the two position variables as part of the class. Second, this implementation only finds the first such overlap. It is possible that there are more than one overlap and later overlaps could be longer. You would have to do a variety of permutations to get the different overlaps (like maybe starting from the back and working forward).

Anyway, good luck with this one.