[hard]57. Insert Interval

57. 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].

Way to easy to be called hard.

class Solution {public:    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {        vector<Interval> ret;                bool inserted = false;        for (auto i: intervals) {            if (i.end < newInterval.start)                ret.push_back(i);            else if (i.start > newInterval.end) {                if (!inserted)                    ret.push_back(newInterval);                ret.push_back(i);                inserted = true;            }            else {                if (i.end >= newInterval.start)                    newInterval.start = min(i.start, newInterval.start);                if (i.start <= newInterval.end)                    newInterval.end = max(newInterval.end, i.end);            }        }        if (!inserted)            ret.push_back(newInterval);        return ret;    }};