57. Insert Interval (Array; Sort)

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].

思路：和上一题差不多，但要考虑更多问题，注意不要漏掉newInterval可能整体插入（不做merge）的情况

/**
 * 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) {
        if(intervals.empty()){
            intervals.push_back(newInterval);
            return intervals;
        }

        //first find the fist end >= newInterval.start
        vector<Interval>::iterator startInsertPos = intervals.begin();
        for(; startInsertPos < intervals.end(); startInsertPos++){
            if(startInsertPos->end >= newInterval.start) break;
        }
        if(startInsertPos == intervals.end()){ //insert in the final
            intervals.push_back(newInterval);
            return intervals;
        }

        //find the last start <= newInterval.end
        vector<Interval>::iterator endInsertPos = startInsertPos;
        for(; endInsertPos < intervals.end(); endInsertPos++){
            if(endInsertPos->start > newInterval.end){
                break;
            }
        }
        endInsertPos--;

        //intervals between [startInsertPos, endInsertPos] may need to be merged
        //case 1: insert before startInsertPos
        if(startInsertPos->start > newInterval.end){
            intervals.insert(startInsertPos,newInterval);//insert in the position startInserPos
        }
        //case2: insert after endInsertPos
        else if(endInsertPos->end < newInterval.start){
            intervals.insert(endInsertPos+,newInterval);//insert in the position endInsertPos+1
        }
        //case3: merge
        else{
            startInsertPos->start = min(newInterval.start, startInsertPos->start);
            startInsertPos->end = max(newInterval.end, endInsertPos->end);
            intervals.erase(startInsertPos+,endInsertPos+);//erase the elem from startInserPos+1 to endInsertPos
        }
        return intervals;