LeetCode #57

# Description:

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:

``````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) {}
* };
*/
``````

# Idea:

compare newInterval和vector里的elements，如果没有overlap，就要么insert，返回，要么往后看。如果有overlap，就更新newInterval，然后删除element。然后再看怎么insert newInterval。

``````std::vector::insert
Return value:
An iterator that points to the first of the newly inserted elements.

std::vector::erase
Return value:
An iterator pointing to the new location of the element that followed the last element erased by the function call.
``````

# Code:

``````class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
auto it=intervals.begin();

while(it != intervals.end()){
if(newInterval.end < it->start){
intervals.insert(it, newInterval);
return intervals;
}
else if(newInterval.start > it->end){
it++;
}
else{
// adjust newInterval, and remove element in intervals
newInterval.start = min(it->start, newInterval.start);
newInterval.end = max(it->end, newInterval.end);
it = intervals.erase(it);
}
}

intervals.insert(intervals.end(), newInterval);
return intervals;
}
};
``````