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/*
leetcode Question 35: Insert Interval
Insert Interval
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].
Analysis:
In other words, the questions gives a new interval, the task is to insert this new interval into an ordered non-overlapping intervals. Consider the merge case.
Idea to solve this problem is quite straight forward:
1. Insert the new interval according to the start value.
2. Scan the whole intervals, merge two intervals if necessary.
*/
/**
* 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 myfunc(Interval a, Interval b) {
return( a.start<b.start);
}
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<Interval> res;
vector<Interval>::iterator it;
for (it = intervals.begin();it!=intervals.end();it++){
if (newInterval.start<(*it).start){
intervals.insert(it,newInterval);
break;
}
}
if (it==intervals.end()){intervals.insert(it,newInterval);}
if (intervals.empty()) {return res;}
res.push_back(*intervals.begin());
for (it = intervals.begin()+1;it!=intervals.end();it++){
if ((*it).start>res.back().end){res.push_back(*it);}
else{ res.back().end = max(res.back().end,(*it).end);}
}
return res;
}
};
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