Day 61, #56, #57, Merge Intervals, Insert Interval

Merge Intervals
Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].
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/**
 * 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:
    static bool comp (const Interval &i1, const Interval &i2) {
        return i1.start < i2.start;
    }

    vector<Interval> merge(vector<Interval> &intervals) {
        sort(intervals.begin(),intervals.end(),comp);
        vector<Interval> ret;
        
        if (intervals.size() == 0) {
            return ret;
        }
        
        ret.push_back(intervals[0]);
        for (int i = 1; i < intervals.size(); i++) {
            if (intervals[i].start <= ret.back().end) {
                if (intervals[i].end > ret.back().end) {
                    ret.back().end = intervals[i].end;
                }
            }else {
                ret.push_back(intervals[i]);
            }
        }
        return ret;
    }
};

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].
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Solution#1

/**
 * 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:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval> ret;
        int index = 0;
        int size = intervals.size();
        
        while (index < size && newInterval.start > intervals[index].end) {
            ret.push_back(intervals[index]);
            index++;
        }
        
        // deal with overlapping
        while (index < size && newInterval.end >= intervals[index].start ) {
            newInterval.start = min(newInterval.start,intervals[index].start);
            newInterval.end = max(newInterval.end,intervals[index].end);
            index++;
        }
        
        ret.push_back(newInterval);
        // push in the rest
        while (index < size) {
            ret.push_back(intervals[index]);
            index++;
        }
        return ret;
    }
};

Solution#2 -- in place
do it later