时间复杂度（斐契那波数求和）

O(2^n)

int fib1(int n) {
  if (n <= 1) return n;
  return fib1(n - 1) + fib1(n - 2);
}

O(n)

int fib2(int n) {
  if (n <= 1) return n;
  int first = 0;
  int second = 1;
  for (int i = 0; i < n - 1; i++) {
    int sum = first + second;
    first = second;
    second = sum;
  }
  return second;
}