Algorithm exercise: number of overlapping intervals

I. Description of the topic

Given a number of possible overlapping intervals, find out the number of overlapping intervals.

Examples are as follows:

Input:[1,5],[ten],[5, Ten],[ +]

Output:2

Note: Test instructions should be to find the maximum number of intervals in the overlap interval, when there is no interval overlap, the maximum number of overlaps is 1, such as

The input is:[1,5],[ 1 ],then the output is ;

Inputs are:[1,2],[2,3],[3, 4] , [4,5] , the output is 2 ( overlapping intervals must intersect with each other );

Enter as:[1,7],[2,5],[3,4],[8,[],[9,+ ],[,), the output is3.

Second, the problem analysis

This problem method is relatively simple, as long as the interval is separated into points, each point has two attributes, one is a value, one is a flag (0 starting point,1 stop), and then the points sorted, and finally, Scan the result of the sorting from the beginning, encounter the starting point overlap number plus 1, encounter the number of stops overlap minus 1, and record the maximum number of overlapping.

The time complexity of this algorithm is O (NLOGN), because the algorithm time is mainly consumed in the sort.

Third, the implementation of the algorithm

#include <iostream> #include <vector> #include <algorithm> #include <cmath> #include < windows.h>using namespace std;//interval definition class interval{public:interval (int iStart, int iend): M_istart (IStart), M_iend ( Iend) {}int m_istart;int m_iend;}; typedef vector<interval> intervalvec;//Interval Split point definition class pointcomparable{public:pointcomparable (int ival, int IType): M_ival (ival), M_itype (iType) {}//overloaded less than operator, sort using bool operator < (const pointcomparable& pcpoint) {if (thi S->m_ival = = pcpoint.m_ival) {return this->m_itype < Pcpoint.m_itype;} return This->m_ival < Pcpoint.m_ival;} int M_ival;int m_itype;//Point type, 0 is the starting point, 1 is the end point};int getoverlappedintervalmaxcount (const intervalvec& intvvec) {vector <PointComparable> pcvec;for (Intervalvec::const_iterator it = Intvvec.begin (); It! = Intvvec.end (); ++it) { Pcvec.push_back (pointcomparable (it->m_istart, 0));p cvec.push_back (Pointcomparable (it->m_iend, 1));} Sort (Pcvec.begin (), Pcvec.end ()), int IMAxcount = 0;int Icurcount = 0;for (Vector<pointcomparable>::iterator ittemp = Pcvec.begin (); ItTemp! = PcVec.end (); ++ittemp) {cout << ittemp->m_ival << "" << ittemp->m_itype << endl;if (ittemp->m_itype = = 0) {Icurcount++;imaxcount = __max (Icurcount, imaxcount);} else{icurcount--;}} return imaxcount;} int main () {Intervalvec intvvec;//intvvec.push_back (Interval (1,5));//Intvvec.push_back (Interval (5,10));// Intvvec.push_back (Interval (1,7));//Intvvec.push_back (Interval (2,5));//Intvvec.push_back (Interval (3,6));// Intvvec.push_back (Interval (8,15));//Intvvec.push_back (Interval (9,17));//Intvvec.push_back (Interval (20,25)); Intvvec.push_back (Interval); Intvvec.push_back (Interval (2,3)); Intvvec.push_back (Interval (3,4)); Intvvec.push_back (Interval (4,5)) cout << "Maximum overlap Interval number:" << getoverlappedintervalmaxcount (Intvvec)  

<span style= "White-space:pre" ></span>cout << Endl;return 0;}

Series Article Description:
1. This series of articles [algorithmic exercises] is just a record of my learning process and self-motivated, and there is no didactic meaning. It is my pleasure to bring a little knowledge and sentiment to the reader.
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Algorithm exercise: number of overlapping intervals