Stanford 算法入门 week 6 Bloom Filter, Hash Function, Search Tree

Bloom Filter 讲解：http://blog.csdn.net/jiaomeng/article/details/1495500


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Programming Question - 6

Question 1

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The goal of this problem is to implement a variant of the 2-SUM algorithm (covered in the Week 5 lecture on hash table applications)

The file contains 500,000 positive integers (there might be some repetitions!).This is your array of integers, with theith 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 [2500,4000] (inclusive) such that there are distinct numbersx,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 1501) in the space provided.

As an optional exercise, you might try implementing your own hash table for this question.

Answer for Question 1

Question 2

Download the text file here.

The goal of this problem is to implement the "Median Maintenance" algorithm (covered in the Week 5 lecture on heap applications). The text file contains a list of the integers from 1 to 10000 in unsorted order; you should treat this as a stream of numbers, arriving one by one. Letting xi denote the ith number of the file, the kth median mk is defined as the median of the numbers x1,…,xk. (So, if k is odd, then mk is ((k+1)/2)th smallest number among x1,…,xk; if k is even, then mk is the (k/2)th smallest number among x1,…,xk.)

In the box below you should type the sum of these 10000 medians, modulo 10000 (i.e., only the last 4 digits). That is, you should compute(m1+m2+m3+⋯+m10000)mod10000.

As an optional exercise, you might compare the performance achieved by heap-based and search-tree-based implementations of the algorithm.

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第一题的代码

#include <iostream>#include <fstream>#define MAX 1000000#define MIN 4000using namespace std;int hash[MIN + 1] = { 0 };int count = 0;void readData() {ifstream fin("HashInt.txt");int temp = 0;while(fin>>temp) {if(temp < MIN) hash[temp]++;}}bool hashMap(int n) {if(n > MIN) return false;if(hash[n]) return true;else return false;}int main() {readData();for(int i = 2500; i <= 4000; i++) {for(int j = 1; j <= (i - 1) / 2; j++) {if(hashMap(j) && hashMap(i - j)) {count++;break;}}}cout<<count<<endl;return 0;}

第二题的思路：http://www.cnblogs.com/lienhua34/archive/2011/12/06/2381299.html

利用大顶堆+小顶堆的方法应该是最高效的
main.cpp的代码， 我的“heap.h”用到的是这里的 http://blog.csdn.net/neostar2008/article/details/7769058

#include <iostream>#include <vector>#include <fstream>#include "heap.h"using namespace std;int main() {ifstream fin("Median.txt");MaxHeap maxh;MinHeap minh;int medianSum = 0;int temp;while(fin>>temp) {if(temp < maxh.top()) maxh.insert(temp);else minh.insert(temp);if(maxh.heapSize() - minh.heapSize() > 1) {minh.insert(maxh.extractMax());}if(minh.heapSize() > maxh.heapSize()) {maxh.insert(minh.extractMin());}medianSum += maxh.top();}cout<<medianSum % 10000<<endl;return 0;}