leetcode93:insert-interval

题目描述

给定一组不重叠的时间区间，在时间区间中插入一个新的时间区间(如果有重叠的话就合并区间)。
这些时间区间初始是根据它们的开始时间排序的。
示例1:
给定时间区间[1,3]，[6,9]，在这两个时间区间中插入时间区间[2,5]，并将它与原有的时间区间合并，变成[1,5],[6,9].

示例2：

给定时间区间[1,2],[3,5],[6,7],[8,10],[12,16],在这些时间区间中插入时间区间[4,9]，并将它与原有的时间区间合并，变成[1,2],[3,10],[12,16].

这是因为时间区间[4,9]覆盖了时间区间[3,5],[6,7],[8,10].

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].

链接：https://www.nowcoder.com/questionTerminal/02418bfb82d64bf39cd76a2902db2190?f=discussion
来源：牛客网

class Solution {

public:

    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {

        vector<Interval> res;

        int i=0;

        for(;i<intervals.size();i++){

            if(newInterval.end<intervals[i].start)

                break;

            else if(newInterval.start>intervals[i].end)

                res.push_back(intervals[i]);

            else{

                newInterval.start=min(newInterval.start,intervals[i].start);

                newInterval.end=max(newInterval.end,intervals[i].end);

            }

          

    }

         

        res.push_back(newInterval);

        for(;i<intervals.size();i++)

            res.push_back(intervals[i]);

    return res;

    }

};