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题目要求：
Download the text file here. (Right click and save link as).
The goal of this problem is to implement a variant of the 2-SUM algorithm (covered in the Week 6 lecture on hash table applications).
The file contains 1 million integers, both positive and negative (there might be some repetitions!).This is your array of integers, with the ith row of the file specifying the ith entry of the array.
Your task is to compute the number of target values t in the interval [-10000,10000] (inclusive) such that there are distinct numbers x,y in the input file that satisfy x+y=t. (NOTE: ensuring distinctness requires a one-line addition to the algorithm from lecture.)
Write your numeric answer (an integer between 0 and 20001) in the space provided.
OPTIONAL CHALLENGE: If this problem is too easy for you, try implementing your own hash table for it. For example, you could compare performance under the chaining and open addressing approaches to resolving collisions.
大致意思是说，有一百万个数字（正负都有，且可能重复）。目标是要寻找到两个数x和y，使得x+y属于[-10000, 10000]的范围中，且x和y在所有数当中都是唯一存在的，最后输出所有可能的和的个数。
文件中的数据差不多是这样子的：
...
6195
2303
5685
1354
4292
7600
6447
4479
9046
7293
5147
1260
1386
6193
4135
3611
8583
...
解题思路：
这道题当然可以采用最暴力的方法，即每次读入一个数后就对数组进行排序，然后记录中位数，但是显然应该还有更好的方法。没错，如果借用“堆”这一数据结构，可以让算法的时间复杂度大大降低。具体的思路如下：
代码实现：
有了上述的思路，利用C++对其进行了实现，代码如下：
#include <iostream> #include <fstream> #include <string> #include <sstream> #include <limits> using namespace std; class MinMaxHeap { public: MinMaxHeap(bool is_min); ~MinMaxHeap(); int Top(); int Size(); void Insert(int num); void Pop(); private: void swap(int index1, int index2); int size; int *element; bool is_min; }; MinMaxHeap::MinMaxHeap(bool is_min = true) { // for this problem, 5010 is just fine this->element = new int[5010]; this->size = 0; this->is_min = is_min; } MinMaxHeap::~MinMaxHeap() { delete[] this->element; } int MinMaxHeap::Top() { return this->element[0]; } int MinMaxHeap::Size() { return this->size; } // The position of each element(the number means the index of the array) // 0 // / // 1 2 // / \ / // 3 4 5 6 void MinMaxHeap::Insert(int num) { int pos = size; element[size++] = num; if (is_min) { while (pos > 0) { int parent = (pos - 1) >> 1; // same as (pos - 1) / 2 if (element[parent] <= element[pos]) { break; } swap(parent, pos); pos = parent; } } else { while (pos > 0) { int parent = (pos - 1) >> 1; // same as (pos - 1) / 2 if (element[parent] >= element[pos]) { break; } swap(parent, pos); pos = parent; } } } void MinMaxHeap::Pop() { element[0] = element[--size]; int pos = 0; if (is_min) { while (pos < (size >> 1)) // if pos >= (size / 2), then element[pos] must be a leaf { int left_child = pos * 2 + 1; int right_child = left_child + 1; int smallest_child; if (right_child < size && element[left_child] > element[right_child]) { smallest_child = right_child; } else { smallest_child = left_child; } if (element[pos] < element[smallest_child]) { break; } swap(pos, smallest_child); pos = smallest_child; } } else { while (pos < (size >> 1)) // if pos >= (size / 2), then element[pos] must be a leaf { int left_child = pos * 2 + 1; int right_child = left_child + 1; int biggest_child; if (right_child < size && element[left_child] < element[right_child]) { biggest_child = right_child; } else { biggest_child = left_child; } if (element[pos] > element[biggest_child]) { break; } swap(pos, biggest_child); pos = biggest_child; } } } void MinMaxHeap::swap(int index1, int index2) { int tmp = element[index1]; element[index1] = element[index2]; element[index2] = tmp; } int main() { ifstream fin; fin.open("Median.txt"); MinMaxHeap MinHeap(true); MinMaxHeap MaxHeap(false); // because we want to find the median, so insert // both min of int and max of int is ok. MinHeap.Insert(numeric_limits<int>::max()); MaxHeap.Insert(numeric_limits<int>::min()); int input, sum = 0, min_top, max_top; string tmp; while (getline(fin, tmp)) { input = atoi(tmp.c_str()); min_top = MinHeap.Top(); max_top = MaxHeap.Top(); if (input < max_top) { MaxHeap.Insert(input); } else { MinHeap.Insert(input); } // balance if (MaxHeap.Size() > MinHeap.Size() + 1) { max_top = MaxHeap.Top(); MaxHeap.Pop(); MinHeap.Insert(max_top); } if (MinHeap.Size() > MaxHeap.Size() + 1) { min_top = MinHeap.Top(); MinHeap.Pop(); MaxHeap.Insert(min_top); } //find the median if (MinHeap.Size() == MaxHeap.Size() + 1) { sum += MinHeap.Top(); } else { sum += MaxHeap.Top(); } } cout << sum % 10000 << endl; fin.close(); system("pause"); return 0; }
通过这个方法，运算效率大大提升。
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原文地址：http://www.cnblogs.com/jdneo/p/4734021.html
