[Leetcode 57, Hard] Insert Interval

Problem:

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:

Solutions:

C++:

vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        if(intervals.empty()) {
            intervals.push_back(newInterval);
            return intervals;
        }

        int start = newInterval.start;
        int end = newInterval.end;
        vector<Interval>::iterator start_it = intervals.begin();
        for(; start_it != intervals.end(); ++start_it) {
            if(start_it->end >= newInterval.start)
                break;
        }
        
        if(start_it == intervals.end()) {
            intervals.push_back(newInterval);
            return intervals;
        } else {
            if(start_it->start > end) {
                intervals.insert(start_it, newInterval);
                return intervals;
            } else if(start_it->end >= end || start_it + 1 == intervals.end() || (start_it + 1)->start > end) {
                start_it->start = min(start, start_it->start);
                start_it->end = max(end, start_it->end);
                return intervals;
            } else
                start_it->start = min(start, start_it->start);
        }
        
        vector<Interval>::iterator end_it = start_it + 1;
        for(; end_it != intervals.end(); ++end_it) {
            if(end_it->end >= end)
                break;
        }
        
        if(end_it == intervals.end())
            start_it->end = end;
        else {
            if(end_it->start <= end) {
                start_it->end = max(end, end_it->end);
                ++end_it;
            } else
                start_it->end = end;
        }
            
        intervals.erase(start_it + 1, end_it);
            
        return intervals;
    }

Java:

Python: