"Leetcode" "Hard" Insert Interval

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

Assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9] , insert and merge in as [2,5] [1,5],[6,9] .

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16] , insert and merge in as [4,9] [1,2],[3,10],[12,16] .

This is because, the new interval [4,9] overlaps with [3,5],[6,7],[8,10] .

Problem Solving Ideas:

1, first use two times to find and to be inserted in the interval has a relationship between the range;

2, will have the relationship of the interval to do processing;

3. Generate a new merged interval and return;

Problem Solving steps:

Code:

1 /**2 * Definition for an interval.3 * struct Interval {4 * int start;5 * int end;6 * Interval (): Start (0), end (0) {}7 * Interval (int s, int e): Start (s), End (e) {}8  * };9  */Ten classSolution { One  Public: A     Static BOOLPartial_order (ConstInterval &a,ConstInterval &b) { -         returnA.end <B.start; -     }; the      -vector<interval> Insert (std::vector<interval> &intervals, Interval newinterval) { -Vector<interval>::iterator less =Lower_bound (Intervals.begin (), Intervals.end (), NewInterval, partial_order); -Vector<interval>::iterator greater =Upper_bound (Intervals.begin (), Intervals.end (), NewInterval, partial_order); +          -Vector<interval>answer; + Answer.insert (Answer.end (), Intervals.begin (), less); A         if(Less <Greater) { atNewinterval.start = Min (Newinterval.start, (*(less ). Start); -Newinterval.end = Max (Newinterval.end, (* (Greater-1) . end); -         } - Answer.push_back (newinterval); - Answer.insert (Answer.end (), Greater, intervals.end ()); -          in         returnanswer; -   } to};

Do not use Lower_bound and upper_bound to write the AC bad code, left for improvement:

1 /**2 * Definition for an interval.3 * struct Interval {4 * int start;5 * int end;6 * Interval (): Start (0), end (0) {}7 * Interval (int s, int e): Start (s), End (e) {}8  * };9  */Ten classSolution { One  Public: Avector<interval> Insert (vector<interval>&intervals, Interval newinterval) { -         if(Intervals.empty ()) -             returnVector<interval> (1, newinterval); the         intleft =0; -         intright = Intervals.size ()-1; -         intSize =intervals.size (); -         intIns_start, Ins_end; +          -          while(Left <Right ) { +             intMid = (left + right)/2; A             if(Intervals[mid].end <Newinterval.start) atLeft = mid +1; -             Else  -right =mid; -         } -Ins_start =Left ; -          inleft = left = =0?0: Left-1; -right = Intervals.size ()-1; to          while(Left <Right ) { +             intMid = (left + right)/2+1; -             if(Newinterval.end <Intervals[mid].start) theright = mid-1; *             Else $left =mid;Panax Notoginseng         } -Ins_end =Right ; the          +         if(Ins_end = =0&& Newinterval.end < intervals[0].start) { A Intervals.insert (Intervals.begin (), newinterval); the             returnintervals; +         } -         if(Ins_start = = Size-1&& Newinterval.start > Intervals[size-1].end) { $ Intervals.push_back (newinterval); $             returnintervals; -         } -          theVector<interval>answer; -Answer.insert (Answer.end (), Intervals.begin (), Intervals.begin () +Ins_start);Wuyi         if(Ins_start <=ins_end) { theNewinterval.start =min (Newinterval.start, intervals[ins_start].start); -Newinterval.end =Max (newinterval.end, intervals[ins_end].end); Wu         } - Answer.push_back (newinterval); AboutAnswer.insert (Answer.end (), Intervals.begin () + Ins_end +1, Intervals.end ()); $         returnanswer; -          -     } -};

"Leetcode" "Hard" Insert Interval