integral time stamps


http://www.1point3acres.com/bbs/thread-177983-1-1.html
interval [startTime, stoptime)   ----integral  time stamps
给这样的一串区间 I1, I2......In  
找出 一个 time stamp  出现在interval的次数最多。
startTime <= t< stopTime 代表这个数在区间里面出现过。

example:  [1,3),  [2, 7),   [4,  8),   [5, 9)
5和6各出现了三次, 所以答案返回5,6。
把所有的startTime和stopTime排序过一遍。碰到start当前interval数+1,否则-1,算出最大的interval overlap数。接下来具体最多的overlap在哪里也很容易找吧
http://blog.5ibc.net/p/52775.html
     vector<int> findTimeStamp(vector<pair<int,int>>& intervals) {
                 vector<int> res;
                 int len=intervals.size();
   if(len==0) return res;
                 vector<int> start(len);
          vector<int> end(len);
   for(pair<int,int> interval:intervals) {
         start.push_back(interval.first);
      end.push_back(interval.second-1);
   }
   sort(start.begin(),start.end());
   sort(end.begin(),end.end());
   int available=0;
   int sIndex=1,eIndex=0;
   int ss=start[0],ee=end[0];
   while(sIndex<len&&eIndex<len) {
         while(start[index]>=end[eIndex]) {
           available++;
        eIndex++;
         }
         if(start[sIndex]<end[eIndex]) {
           if(available<=0) {
      ss=start[sIndex];  //记录起点和终点
      ee=end[eIndex];
        }
        else {
             available--;
        }
        }
               sIndex++;
   }
   for(int i=ss;i<=ee;i++) {
       res.push_back(i);
   }
   return res;
  } 

这题就是LC上面的meeting rooms换了马甲。meeting rooms求最少需要多少个room,也就是某个时刻有几个并行的会议,这个题目求时刻有最大的并行interval数量,差不多,按照start time排一遍,再用一个heap放结束时间就可以了
  1. def integraltimestamps2(intervals):.鏈枃鍘熷垱鑷�1point3acres璁哄潧
  2.     intvs = sorted(intervals, key = lambda x : x[0])
  3.     edges = sorted([intv[0] for intv in intervals]+[intv[1] for intv in intervals])
  4.     results = []. visit 1point3acres.com for more.
  5.     moccurs = 0
  6.     ongoing = []
  7.     intvid = 0
  8.     for e in edges:. 鍥磋鎴戜滑@1point 3 acres
  9.         if results and len(ongoing) == moccurs:
  10.             results += range(results[-1]+1,e)
  11.         while ongoing and e >= ongoing[0]:
  12.             heapq.heappop(ongoing). 1point 3acres 璁哄潧
  13.         while intvid < len(intvs) and e >= intvs[intvid][0]:
  14.             heapq.heappush(ongoing, intvs[intvid][1]). from: 1point3acres.com/bbs 
  15.             intvid += 1
  16.         if len(ongoing) > moccurs:
  17.             moccurs = len(ongoing)
  18.             results = [e]
  19.         elif len(ongoing) == moccurs:.鏈枃鍘熷垱鑷�1point3acres璁哄潧
  20.             results.append(e)
  21.     return results



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