I would find out who had the time to put a billion numbers into an array and fire him. Must work for government. At least if you had a linked list you could insert a number into the middle without moving half a billion to make room. Even better a Btree allows for a binary search. Each comparison eliminates half of your total. A hash algorithm would allow you to populate the data structure like a checkerboard but not so good for sparse data. As it is your best bet is to have a solution array of 100 integers and keep track of the lowest number in your solution array so you can replace it when you come across a higher number in the original array. You would have to look at every element in the original array assuming it is not sorted to begin with.

您可以使用快速选择算法在(按顺序)索引[十亿-101]处查找数字 然后遍历这些数字找出比这个数字更大的数。

array={...the billion numbers...} 
result[100];

pivot=QuickSelect(array,billion-101);//O(N)

for(i=0;i<billion;i++)//O(N)
   if(array[i]>=pivot)
      result.add(array[i]);

该算法时间为:2 X O(N) = O(N)(平均情况性能)

Thomas Jungblut建议的第二个选择是:

使用堆构建最大堆将花费O(N)，然后前100个最大的数字将在堆的顶部，所有你需要的是把它们从堆(100 X O(Log(N))。

该算法时间为:O(N) + 100 X O(Log(N)) = O(N)

我看到了很多O(N)的讨论，所以我提出了一些不同的想法。

关于这些数字的性质有什么已知的信息吗?如果答案是随机的，那就不要再进一步了，看看其他答案。你不会得到比他们更好的结果。

However! See if whatever list-populating mechanism populated that list in a particular order. Are they in a well-defined pattern where you can know with certainty that the largest magnitude of numbers will be found in a certain region of the list or on a certain interval? There may be a pattern to it. If that is so, for example if they are guaranteed to be in some sort of normal distribution with the characteristic hump in the middle, always have repeating upward trends among defined subsets, have a prolonged spike at some time T in the middle of the data set like perhaps an incidence of insider trading or equipment failure, or maybe just have a "spike" every Nth number as in analysis of forces after a catastrophe, you can reduce the number of records you have to check significantly.

不管怎样，还是有一些值得思考的东西。也许这会帮助你给未来的面试官一个深思熟虑的回答。我知道，如果有人问我这样一个问题来回应这样的问题，我会印象深刻——这将告诉我，他们正在考虑优化。只是要认识到，优化的可能性并不总是存在的。

使用第n个元素得到第100个元素O(n) 迭代第二次，但只有一次，并输出大于此特定元素的所有元素。

请特别注意，第二步可能很容易并行计算!当你需要一百万个最大的元素时，它也会很有效。

我对此的直接反应是使用堆，但有一种方法可以使用QuickSelect，而不需要在任何时候保留所有的输入值。

创建一个大小为200的数组，并用前200个输入值填充它。运行QuickSelect并丢弃低100个位置，留下100个空闲位置。读入接下来的100个输入值并再次运行QuickSelect。继续执行，直到以100个批次为单位运行整个输入。

最后是前100个值。对于N个值，您运行QuickSelect大约N/100次。每个快速选择的代价大约是某个常数的200倍，所以总代价是某个常数的2N倍。在我看来，输入的大小是线性的，不管我在这个解释中硬连接的参数大小是100。

两个选择:

(1)堆(priorityQueue)

维护最小堆的大小为100。遍历数组。一旦元素小于堆中的第一个元素，就替换它。

InSERT ELEMENT INTO HEAP: O（log100）
compare the first element: O(1)
There are n elements in the array, so the total would be O(nlog100), which is O(n)

(2)映射-约简模型。

这与hadoop中的单词计数示例非常相似。 映射工作:计算每个元素出现的频率或次数。 减约:获取顶部K元素。

通常，我会给招聘人员两个答案。他们喜欢什么就给什么。当然，映射缩减编码会很费事，因为您必须知道每个确切的参数。练习一下也无妨。 祝你好运。