57. Insert Interval

题目

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].

解析

题意：给了一个区间数组与一个区间，将区间合入区间数组中
解析：先将给的单独区间合入数组，然后如同56题那样合并区间即可

代码

时间复杂度：O(n)
先附上自己的：

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

下面是discuss中简短的：

vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
    vector<Interval> ret;
    auto it = intervals.begin();
    for(; it!=intervals.end(); ++it){
        if(newInterval.end < (*it).start) //all intervals after will not overlap with the newInterval
            break; 
        else if(newInterval.start > (*it).end) //*it will not overlap with the newInterval
            ret.push_back(*it); 
        else{ //update newInterval bacause *it overlap with the newInterval
            newInterval.start = min(newInterval.start, (*it).start);
            newInterval.end = max(newInterval.end, (*it).end);
        }   
    }
    // don't forget the rest of the intervals and the newInterval
    ret.push_back(newInterval);
    for(; it!=intervals.end(); ++it)
        ret.push_back(*it);
    return ret;
}