LeetCode--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].

思路：这一题是上一题的加强版，我延续了上一题的想法。首先判断是否可以插入并找到插入的位置，如果可以插入，则合并已经放入的最后一个区间和插入的区间，再把后面的区间和这个合并后的区间继续合并，和上一题一样；如果不可以则说明有两种特殊情况，该区间在第一个或者是最后一个，插入后排序即可。

/** * 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>result;        if(intervals.empty()){            result.push_back(newInterval);            return result;        }        int index=0;        bool flag=false;        for(int i=0;i<intervals.size();i++){            result.push_back(intervals[i]);            if(newInterval.start<=intervals[i].end&&newInterval.end>=intervals[i].start){                result.erase(result.end());                intervals[i].start=min(newInterval.start,intervals[i].start);                intervals[i].end=max(newInterval.end,intervals[i].end);                index=i;                flag=true;                result.push_back(intervals[i]);                break;            }        }        if(flag){            for(int i=index+1;i<intervals.size();i++){                if(intervals[i].start<=result.back().end){                    result.back().end=max(intervals[i].end,result.back().end);                }                else{                    result.push_back(intervals[i]);                }            }        }        else{            result.push_back(newInterval);            sort(result.begin(),result.end(),[](Interval a,Interval b){return a.start<b.start;});        }        return result;    }};