c++ - 算法时间复杂度计算

Question

请问有人知道下面算法的时间复杂度是多少吗？，如果知道，希望您能及时的帮一下忙。如果没有if 语句，单纯的两个嵌套循环，可以知道时间 复杂度为O(n^2),
for (i = 0; i < num; i++)
{
for (int j = 0; j < num; j++)
if (dataItem[i] == dataItem[j])
{
cou++;
if (cou > num / 2)
break;
}
if (cou > num / 2)
break;
}

Answer 1

O(n^2)
You can think of Big O Notation as representing the worst-case time complexity.
In the worst case, all dataItems are not equal, and the loop will be executed approximately n^2/2 + n/2 times.