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].
/** * 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) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<Interval> result; int i = 0; int n = intervals.size(); while( i < n && newInterval.start > intervals[i].end ) result.push_back(intervals[i++]); while(i < n && newInterval.end >= intervals[i].start) { newInterval.start = newInterval.start < intervals[i].start ? newInterval.start : intervals[i].start; newInterval.end = newInterval.end > intervals[i].end ? newInterval.end : intervals[i].end; i++; } result.push_back(newInterval); while(i< n) result.push_back(intervals[i++]) ; return result ; } };