采访Q：数组的查找方式[重复]

Question

我最近在面试，面试官问了我以下问题：

Given an unsorted array, how do you calculate the mode in O(N)?

我的回答是使用哈希图、O(N) 遍历数组和 O(1) 查找。

然后他说

If you had to use constant memory but were allowed more processor time, how would you do it?

我回答'对数组排序并找到最长的运行时间，runtime = O(nlgn)

他问的下一个问题把我搞砸了..

If you had to use constant memory and linear time how would you do it?

我不知道如何回答这个问题，他把这个留给我作为以后的练习。已经好几天了，我还是不知道该怎么做。

谁能知道怎么做？>

Answer 1

如果数字在合理范围内，您可以使用具有值范围大小的数组线性计算众数。

数组中的值将是 mode 数组的索引。增加值。保留其他两个频率最高的变量和频率最高的值的索引。

#include <stdio>

int main(void)
{
  unsigned int frequencies[11] = {0}; // Assume range 1..10, inclusive.
  const unsigned int values[] =
    {1, 8, 4, 8, 7, 3, 2, 8, 5};
  const unsigned int value_quantity =  
    sizeof(values) / sizeof(values[0]);
  unsigned int greatest_frequency = 0;
  unsigned int value_of_greatest_frequency = 0;

  for (unsigned int i = 0; i < value_quantity; ++i)
  {
    // Calculate new frequency.
    const unsigned int frequency_index = values[i];
    ++frequencies[frequency_index];

    // Update "running" variables
    if (frequencies[frequency_index] > greatest_frequency)
    {
      greatest_frequency = frequencies[frequency_index];
      value_of_greatest_frequency = frequency_index;
    }
  }
  std::cout << "Mode is " << value_of_greatest_frequency
            << ", with frequency of " << greatest_frequency
            << "\n";
  return 0;
}

一次通过，许多变量。这只有在值的范围是合理的情况下才有效。

另一个建议是使用std::map</*value*/, /*frequency*/>。

这个算法不是常数时间，而是O(N)或者更少。