为什么Big-O是O（N）而不是此代码的O（N/2）

发布于 2025-01-28 05:14:44 字数 704 浏览 1 评论 0原文

我试图了解算法的大o。

我在网上找到了此代码，但我无法理解如何为其计算Big-O：

void printFirstItemThenFirstHalfThenSayHi100Times(int arr[], int size)
{
    printf("First element of array = %d\n",arr[0]); % O(1)
    
    for (int i = 0; i < size/2; i++)                % O(n/2)
    {
        printf("%d\n", arr[i]);
    }

    for (int i = 0; i < 100; i++)                   % O(100), which is a constant
    {
        printf("Hi\n");
    }
}

它说（ big-o-with-示例），big-o是o（n）！为什么？

我看到我们在开始时有一个常数O（1），然后是O（N/2），最后是O（100）。为什么big-o是o（n）而不是o（n/2）？

原文

I am trying to understand the Big-O of algorithms.

I found this code online and I wasn't able to understand how the Big-O was calculated for it:

void printFirstItemThenFirstHalfThenSayHi100Times(int arr[], int size)
{
    printf("First element of array = %d\n",arr[0]); % O(1)
    
    for (int i = 0; i < size/2; i++)                % O(n/2)
    {
        printf("%d\n", arr[i]);
    }

    for (int i = 0; i < 100; i++)                   % O(100), which is a constant
    {
        printf("Hi\n");
    }
}

It says (big-o-with-examples) that, the big-O is O(N)! why?

I see that we have a constant at the beginning O(1), then O(n/2), and finally O(100). Why the Big-O is O(N) and not O(N/2)?

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