"Leetcode" Insert Interval

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] .

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) { -         -         intn=intervals.size (); theVector<interval>result; -         -         if(n==0) -         { + Result.push_back (newinterval); -             returnresult; +         } A         at         intstart=Newinterval.start; -         intEnd=Newinterval.end; -   -         -         BOOLisfindstart=false; -         BOOLIsfindend=false; in         BOOLAddcur=true; -         to          for(intI=0; i<n;i++) +         { -Addcur=true; the             *             if(!Isfindstart) $             {Panax Notoginseng                 //does not intersect with the current interval and is in front of start -                 if(start>intervals[i].end) the                 { +                     //need to add current element AAddcur=true; the                     //when the current element is the last element, it is assumed that start has found the +                     if(i==n-1) isfindstart=true; -                 } $                 Else $                 { -                     //intersect with current interval or start less than current interval -                     thestart=min (start,intervals[i].start); -                     //you don't have to start looking for the next.Wuyiisfindstart=true; the                     -                     //if start and end do not have a cross with other interval Wu                     if(End<intervals[i].start) addcur=true; -                     ElseAddcur=false; About                 } $             } -             -             -             if(!isfindend) A             { +                 //need to continue looking for end the                 if(end>intervals[i].end) -                 { $                     the                     if(start<=intervals[i].end) addcur=false; the   the                     if(i==n-1) the                     { -Isfindend=true; in                         the                         if(addcur) result.push_back (Intervals[i]); the Result.push_back (Interval (start,end)); About                         the                         Continue; the                     } the                 } +                 Else -                 { the                     //end is less than the current interval and the end is found at this timeBayiEnd=end>=intervals[i].start?Intervals[i].end:end; the   the                     //found the end -Isfindend=true; -                     if(End>=intervals[i].start) addcur=false; the                     the Result.push_back (Interval (start,end)); the                     if(addcur) result.push_back (Intervals[i]); the                     -                     Continue; the                 } the             } the            94             if(addcur) result.push_back (Intervals[i]); the         } the         returnresult; the     }98};

"Leetcode" Insert Interval