LeetCode 253. Meeting Rooms II

最新推荐文章于 2024-08-07 09:52:27 发布

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本文介绍了一种高效的算法来解决会议室调度问题。通过将会议室预约的时间段转换为一系列时间点，并使用排序和计数策略来确定所需的最小会议室数量。此外，还探讨了如何找到最佳的会议时间点。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

/*
  Given an array of meeting time interval consiting of start and end times
  [[s1, e1], [s2, e2], [s3, e3]] (si < ei), find the mininum number of
  conference rooms required.
  For example:
  Given [[0, 30], [5, 10], [15, 20]]
  return 2.
*/
struct Interval {
  int start;
  int end;
  Interval(int a, int b) : start(a), end(b) {}
};

static bool sorter(int a, int b) {
  return abs(a) < abs(b);
}

int minMeetingRooms(vector<Interval>& intervals) {
  int res = 0;
  if(intervals.size() == 0) return res;
  if(intervals.size() == 1) return res + 1;
  vector<int> times;
  for(int i = 0; i < intervals.size(); ++i) {
    times.push_back(intervals[i].start);
    times.push_back(-intervals[i].end);
  }
  sort(times.begin(), times.end(), sorter);
  int count = 0;
  for(int i = 0; i < times.size(); ++i) {
    if(times[i] >= 0) count++;
    else count--;
    res = max(res, count);
  }
  return res;
}
int main(void) {
  Interval time_1(0, 30);
  Interval time_2(5, 10);
  Interval time_3(15, 20);
  vector<Interval> intervals;
  intervals.push_back(time_1);
  intervals.push_back(time_2);
  intervals.push_back(time_3);
  cout << minMeetingRooms(intervals) << endl;
}

Another variation to find the best meeting time. Best meeting time is absolutely the maxNumber of houses required if the conference runs in parallel.

#include "header.h"
using namespace std;

struct Interval{
  int start;
  int end;
  Interval(int s, int e) : start(s), end(e) {}
};

static inline bool sorter(int a, int b) {
  return abs(a) < abs(b);
}

int bestMeetingPoint(vector<Interval>& intervals) {
  vector<int> times;
  for(int i = 0; i < intervals.size(); ++i) {
    times.push_back(intervals[i].start);
    times.push_back(-intervals[i].end);
  }
  int maxCount = 0;
  int bestTime = 0;
  sort(times.begin(), times.end(), sorter);
  int count = 0;
  for(int i = 0; i < times.size(); ++i) {
    if(times[i] >= 0) count++;
    else count--;
    if(maxCount < count) {
      maxCount = count;
      bestTime = times[i];
    }
  }
  return bestTime;
}

int main(void) {
  vector<Interval> intervals{Interval{1, 3}, Interval{2, 7}, Interval{4, 8}, Interval{5, 9}};
  cout << bestMeetingPoint(intervals) << endl;
}