计算大数组中的唯一元素 [英] Counting unique element in large array

问题描述

在采访中，我的一位同事在提问后被问到。

One of my colleague was asked following question in an interview.

给出一个存储未签名int的巨大数组。数组的长度为100000000。找到计算数组中元素的唯一数量的有效方法。


Eg arr = {2,34,5,6,7 ，2,2,5,1,34,5}
O / p：2的计数为3，34的计数为2，依此类推。

Given a huge array which stores unsigned int. Length of array is 100000000. Find the effective way to count the unique number of elements present in the array.
E.g arr = {2,34,5,6,7,2,2,5,1,34,5} O/p: Count of 2 is 3, Count of 34 is 2 and so on.

有效的算法是什么？我以为首先字典/哈希是一种选择，但是由于数组很大，因此效率很低。

What is the effective algorithms to do this? I thought at first dictionary/hash would be one of the option, but since array is very large it is in-efficient. Is there any way to do this?

谢谢，
chota

Thanks, chota

推荐答案

堆排序为O（nlogn）并就地进行。处理大型数据集时，就地是必要的。排序后，您可以遍历数组，计算每个值的出现次数。因为数组是排序的，所以一旦值发生变化，您就会知道所有先前值的出现。

Heap sort is O(nlogn) and in-place. In-place is necessary when dealing with large data sets. Once sorted you can make one pass through the array tallying occurrences of each value. Because the array is sorted, once a value changes you know you've seen all occurrences of the previous value.

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