Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9]
, insert and merge [2,5]
in as [1,5],[6,9]
.
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16]
, insert and merge [4,9]
in as [1,2],[3,10],[12,16]
.
This is because the new interval [4,9]
overlaps with [3,5],[6,7],[8,10]
.
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */ class Solution { public: bool compare(Interval& i1,Interval&i2){ if(i1.start<i2.start)return true; else if(i1.start==i2.start){ return i1.end<i2.end; } else return false; } vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { vector<Interval> ret; bool put=false; for(int i=0;i<intervals.size();i++){ if(intervals[i].end<newInterval.start||intervals[i].start>newInterval.end){ if(!put&&intervals[i].start>newInterval.end){ ret.push_back(newInterval); put=true; } ret.push_back(intervals[i]); } else{ newInterval.start=min(newInterval.start,intervals[i].start); newInterval.end=max(newInterval.end,intervals[i].end); } } if(!put)ret.push_back(newInterval); return ret; } };