时间复杂度频率表实例

例1

语句                                                执行步数        频率         总步数

int factorial(int n)                                   0             0           Theta(0)
{                                                      0             0           Theta(0)
   if (n <= 1) return 1;                               1             1           Theta(1)
   else return n * factorial(n - 1);   1 + tfactorial(n-1)             1   1 + tfactorial(n-1)
}                                                      0             0           Theta(0)

factorial是计算n！的递归函数，所以

当n=1是设执行factorial时间是c， 我们可以得到：

t_factorial(n) = 1 + t_factorial(n-1) = 2 + t_factorial(n-2) = 3 + t_factorial(n-3)
...... = n - 1 + t_factorial(1)= n - 1 + c = Theta(n)

所以 factorial得时间复杂度是 Theta(n)

例2

_______________________________________________________________________________________
语句                                            执行步数            频率       总步数
_______________________________________________________________________________________
bool minmax(...)                                    0                0         Theta(0)
{                                                   0                0         Theta(0)
   if (n < 1) return false;                         1                1         Theta(1)
   indexOfMin = indexOfMax = 0;                     1                1         Theta(1)
   for (int i = 1; i < n; i++)                      1         Theta(n)         Theta(n)
      if (a[indexOfMin] > a[i]) indexOfMin = i;     1         Theta(n)         Theta(n)
      else if (a[indexOfMax] ...) indexOfMax = i;   1   Omega(0), O(n)   Omega(0), O(n)
   return true;                                     1                1         Theta(1)
}                                                   0                0         Theta(0)
_______________________________________________________________________________________

因此, t_minmax(n) = Theta(n), n >= 1.