最好和最坏的情况

Question

I need help to find best case and worst case of this code with explanation. I think worst case is O(n).

public static boolean adjacentDuplicates(int[] a) {
  for (int i = 0; i < a.length-1; i++)
    if (a[i] == a[i+1]) return true;
  return false;
}

Answer 1

The best case is in the first comparison return the value. Hence, if a[0] == a[1] the time complexity is \\Theta(1) . And the worse is the comparison will not be satisfied up to the end of the loop. Hence, the worst-case complexity is \\Theta(n) (that n is the length of the input array a ).