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c++ - 在 1.5 秒内找到超过 2000 万个 3 到 4 个不同整数的中位数

问题描述

我正在尝试排序并找到仅包含 3 到 4 个不同整数的整数字符串的中位数。

我正在处理的数字数量约为 20 到 2500 万，我应该对向量进行排序并在每次将新整数添加到向量中时找到中值并将中值添加到单独的“总计”变量中每次生成中位数时，它会汇总所有中位数。

1                   Median: 1              Total: 1
1 , 2               Median: (1+2)/2 = 1    Total: 1 + 1 = 2
1 , 2 , 3           Median: 2              Total: 2 + 2 = 4
1 , 1 , 2 , 3       Median: (1+2)/2 = 1    Total: 4 + 1 = 5
1 , 1 , 1 , 2 , 3   Median: 1              Total: 5 + 1 = 6

我正在尝试找到一种方法来进一步优化我的代码，因为它不够高效。（必须在 2 秒左右的时间内处理）有谁知道如何进一步加快我的代码逻辑？

我目前在 C++ 中使用 2 个堆或优先级队列。一个用作最大堆，另一个用作最小堆。

从数据结构中得到了找到中位数的想法

You can use 2 heaps, that we will call Left and Right.
Left is a Max-Heap.
Right is a Min-Heap.
Insertion is done like this:

If the new element x is smaller than the root of Left then we insert x to 
Left.
Else we insert x to Right.
If after insertion Left has count of elements that is greater than 1 from 
the count of elements of Right, then we call Extract-Max on Left and insert 
it to Right.
Else if after insertion Right has count of elements that is greater than the 
count of elements of Left, then we call Extract-Min on Right and insert it 
to Left.
The median is always the root of Left.

So insertion is done in O(lg n) time and getting the median is done in O(1) 
time.

但这还不够快...

标签： c++performancedata-structuresbig-omedian

解决方案

如果您在字符串中只有三到四个不同的整数，您可以通过遍历字符串一次来跟踪每个整数出现的次数。从此表示中添加（和删除元素）也可以在恒定时间内完成。

class MedianFinder
{
public:
  MedianFinder(const std::vector<int>& inputString)
  {
    for (int element : inputString)
      _counts[element]++; // Inserts 0 into map if element is not in there.
  }

  void addStringEntry(int entry)
  {
    _counts[entry]++;
  }

  int getMedian() const
  {
    size_t numberOfElements = 0;
    for (auto kvp : _counts)
      numberOfElements += kvp.second;

    size_t cumulativeCount = 0;
    int lastValueBeforeMedian;
    for (auto kvp : _counts)
    {
      cumulativeCount += kvp.second;
      if (cumulativeCount >= numberOfElements/2)
        lastValueBeforeMedian = kvp.first;
    }

    // TODO! Handle the case of the median being in between two buckets.
    //return ...
  }

private:
  std::map<int, size_t> _counts;
};

这里没有显示对中位数求和的琐碎任务。

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