leetcode-57 insert interval

有了上道题的基础，这道题将新来的interval插入到原来的intervals当中，然后进行融合即可

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].

class Solution {public:static bool comp(const Interval &a, const Interval &b){return a.start < b.start;}vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {vector<Interval> res;if (intervals.empty()){res.push_back(newInterval);return res;}intervals.push_back(newInterval);sort(intervals.begin(), intervals.end(), comp);mergeIntervals(intervals, res);return res;}void mergeIntervals(vector<Interval> &intervals, vector<Interval> &res){int n = intervals.size();int s = intervals[0].start;int tempend = intervals[0].end;int i = 1;while (i < n){if (intervals[i].start <= tempend){tempend = max(tempend, intervals[i].end); //若有重合，判断集合右边谁比较大i++;}else{Interval temp = Interval(s, tempend);res.push_back(temp);s = intervals[i].start;tempend = max(intervals[i].end, tempend);i++;}}tempend = max(tempend, intervals[i - 1].end);Interval temp = Interval(s, tempend);res.push_back(temp);}};