// Source : https://oj.leetcode.com/problems/insert-interval/
// Author : Hao Chen
// Date : 2014-08-26
/**********************************************************************************
*
* 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].
*
*
**********************************************************************************/
#include
#include
#include
using namespace std;
struct Interval {
int start;
int end;
Interval() : start(0), end(0) {}
Interval(int s, int e) : start(s), end(e) {}
};
bool compare(const Interval& lhs, const Interval& rhs){
return (lhs.start==rhs.start) ? lhs.end < rhs.end : lhs.start < rhs.start;
}
vector merge(vector &intervals) {
vector result;
if (intervals.size() <= 0) return result;
sort(intervals.begin(), intervals.end(), compare);
for(int i=0; i0 && result[size-1].end >= intervals[i].start) {
result[size-1].end = max(result[size-1].end, intervals[i].end);
}else{
result.push_back(intervals[i]);
}
}
return result;
}
vector insert(vector &intervals, Interval newInterval) {
intervals.push_back(newInterval);
return merge(intervals);
}
int main(int argc, char**argv)
{
Interval i1(1,2);
Interval i2(3,5);
Interval i3(6,7);
Interval i4(8,10);
Interval i5(12,16);
vector intervals;
intervals.push_back(i1);
intervals.push_back(i2);
intervals.push_back(i3);
intervals.push_back(i4);
intervals.push_back(i5);
Interval n(4,9);
vector r = insert(intervals, n);
for(int i=0; i