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insert_interval.cpp
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56 lines (45 loc) · 1.59 KB
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/*
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) {
vector<Interval> res;
for (int i = 0; i < intervals.size(); i++) {
int s1 = intervals[i].start;
int e1 = intervals[i].end;
int s2 = newInterval.start;
int e2 = newInterval.end;
if ((s1 <= s2 && s2 <= e1) || (s2 <= s1 && s1 <= e2)) { // embedded
newInterval.start = min(s1, s2);
newInterval.end = max(e1, e2);
} else if (e1 < s2) {
res.push_back(intervals[i]);
} else if (e2 < s1) {
res.push_back(newInterval);
for (int j = i; j < intervals.size(); j++)
res.push_back(intervals[j]);
return res;
} else {
cout << "should not happen";
}
}
res.push_back(newInterval);
return res;
}
};