-
Notifications
You must be signed in to change notification settings - Fork 0
/
Insert-Interval.cpp
60 lines (44 loc) · 1.47 KB
/
Insert-Interval.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
/*
* Song Qiang <qiang.song@usc.edu> 2013
*/
/*
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;
while (i < intervals.size() && intervals[i].end < newInterval.start)
result.push_back(intervals[i++]);
while (i < intervals.size() && intervals[i].start <= newInterval.end)
{
newInterval.start = min(intervals[i].start, newInterval.start);
newInterval.end = max(intervals[i].end, newInterval.end);
++i;
}
result.push_back(newInterval);
while (i < intervals.size()) result.push_back(intervals[i++]);
return result;
}
};