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41_StreamMedian.cpp
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144 lines (114 loc) · 3.5 KB
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/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题41：数据流中的中位数
// 题目：如何得到一个数据流中的中位数？如果从数据流中读出奇数个数值，那么
// 中位数就是所有数值排序之后位于中间的数值。如果从数据流中读出偶数个数值，
// 那么中位数就是所有数值排序之后中间两个数的平均值。
#include <cstdio>
#include <algorithm>
#include <vector>
#include <functional>
using namespace std;
template<typename T> class DynamicArray
{
public:
void Insert(T num)
{
if(((min.size() + max.size()) & 1) == 0) //数据的总数目是偶数时把新数据插入最小堆 （小顶堆）
{
if(max.size() > 0 && num < max[0]) //新数据比最大堆（大顶堆）中的一些数据要小
{
max.push_back(num);
push_heap(max.begin(), max.end(), less<T>());
num = max[0];
pop_heap(max.begin(), max.end(), less<T>());
max.pop_back();
}
min.push_back(num);
push_heap(min.begin(), min.end(), greater<T>());
}
else //数据的总数目是奇数时把新数据插入最大堆 （大顶堆）
{
if(min.size() > 0 && min[0] < num) //新数据大于最小堆（小顶堆）中的一些数据
{
min.push_back(num);
push_heap(min.begin(), min.end(), greater<T>());
num = min[0];
pop_heap(min.begin(), min.end(), greater<T>());
min.pop_back();
}
max.push_back(num);
push_heap(max.begin(), max.end(), less<T>());
}
}
T GetMedian()
{
int size = min.size() + max.size();
if(size == 0)
{
//throw exception("No numbers are available");
throw logic_error("No numbers are available");
}
T median = 0;
if((size & 1) == 1)
median = min[0];
else
median = (min[0] + max[0]) / 2;
return median;
}
private:
vector<T> min;
vector<T> max;
};
// ====================测试代码====================
void Test(char* testName, DynamicArray<double>& numbers, double expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
if(abs(numbers.GetMedian() - expected) < 0.0000001)
printf("Passed.\n");
else
printf("FAILED.\n");
}
int main(int argc, char* argv[])
{
DynamicArray<double> numbers;
printf("Test1 begins: ");
try
{
numbers.GetMedian();
printf("FAILED.\n");
}
catch(const exception&)
{
printf("Passed.\n");
}
numbers.Insert(5);
Test("Test2", numbers, 5);
numbers.Insert(2);
Test("Test3", numbers, 3.5);
numbers.Insert(3);
Test("Test4", numbers, 3);
numbers.Insert(4);
Test("Test6", numbers, 3.5);
numbers.Insert(1);
Test("Test5", numbers, 3);
numbers.Insert(6);
Test("Test7", numbers, 3.5);
numbers.Insert(7);
Test("Test8", numbers, 4);
numbers.Insert(0);
Test("Test9", numbers, 3.5);
numbers.Insert(8);
Test("Test10", numbers, 4);
return 0;
}