在Java中的整数数组中找到模式长度？

Question

Suppose I have a Fibonacci sequence of the number x as follows and I want to detect a sequence in an array. Java method should return the length of sequence

   x         0  1  1  2  3  5  8  13  21  34  55  89  144  233  377  610
 1)x mod 2 - 0  1  1  0  1  1  0   1   1   0   1   1    0    1    1    0
 2)x mod 3 - 0  1  1  2  0  2  2   1   0   1   1   2    0    2    2    1

 Answer 1) 3 (repetitive sequence 011 and length is 3)
        2) 8 (repetitive sequence 01120221 and length is 8)

Answer 1

A very simple approach would be to create a copy of the array, and check position 0 of the first array against position 1 of the second array, and if they match, continue checking until the end. If the whole thing matches, then you have a repeating length of 1.

If not, then you compare position 0 of the first array to position 2 of the second array, and follow the same process as above. If this matches, then you have a repeating length of 2.

And you continue this process until you either find a match, or reach the end of the array and can't offset it any further, in which case there is no repeat.

Answer 2

If you are only intending to use this specifically for modulo values of numbers in the Fibonacci sequence, and not for arbitrary data, then the sequence will repeat as soon as you find the second occurrence of a 0 followed by a 1 in the modulo sequence. This is because mod(a + b, n) = mod(mod(a, n) + mod(b, n), n), so the modulo of a number in the Fibonacci sequence (which is the sum of the two previous values) is determined by the previous 2 modulo results. Therefore, once the original pattern of a 0 followed by a 1 reoccurs, the pattern will repeat.

Answer 3

The Following code works:

    private static int detectSequence(int[] array) {

    int count = 2;
    for (int i = 2; i < array.length; i++) {
        if(array[i] == 0 && array[i+1] == 1 && i+1 < array.length){
            return count;
        }
        count++;
    }
    return count;

}