lkeetcode 57. 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].

class Solution {public:vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {if (intervals.empty()){intervals.push_back(newInterval);return intervals;}if (newInterval.end < intervals[0].start){intervals.insert(intervals.begin(), newInterval);return intervals;}if (newInterval.end == intervals[0].start){intervals[0].start= newInterval.start;return intervals;}int k = 0;while (k <intervals.size()){while (k <intervals.size() && intervals[k].end < newInterval.start)k++;if (k == intervals.size()){intervals.push_back(newInterval);return intervals;}if (intervals[k].start > newInterval.end){intervals.insert(intervals.begin() + k, newInterval);return intervals;}intervals[k].start = newInterval.start < intervals[k].start ? newInterval.start: intervals[k].start;if (intervals[k].end >= newInterval.end)return intervals;int end = newInterval.end;int kk = k;while (kk < intervals.size() && end >= intervals[kk].end)kk++;if (kk == intervals.size()){intervals[k].end = end;intervals.erase(intervals.begin() + k + 1, intervals.end());return intervals;}if (intervals[kk].start <= end){intervals[k].end = intervals[kk].end;intervals.erase(intervals.begin() + k + 1, intervals.begin() + kk+1);return intervals;}intervals[k].end = end;intervals.erase(intervals.begin() + k + 1, intervals.begin() + kk);return intervals;}}};

accepted