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].
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{% if book.java %}
struct Interval {
int start;
int end;
Interval() : start(0), end(0) { }
Interval(int s, int e) : start(s), end(e) { }
};
// Insert Interval
// 时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
for (int i = 0; i < intervals.size();) {
final Interval cur = intervals.get(i);
if (newInterval.end < cur.start) {
intervals.add(i, newInterval);
return intervals;
} else if (newInterval.start > cur.end) {
++i;
continue;
} else {
newInterval.start = Math.min(newInterval.start, cur.start);
newInterval.end = Math.max(newInterval.end, cur.end);
intervals.remove(i);
}
}
intervals.add(newInterval);
return intervals;
}
}{% endif %}
{% if book.cpp %}
// Insert Interval
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval>::iterator it = intervals.begin();
while (it != intervals.end()) {
if (newInterval.end < it->start) {
intervals.insert(it, newInterval);
return intervals;
} else if (newInterval.start > it->end) {
it++;
continue;
} else {
newInterval.start = min(newInterval.start, it->start);
newInterval.end = max(newInterval.end, it->end);
it = intervals.erase(it);
}
}
intervals.insert(intervals.end(), newInterval);
return intervals;
}
};{% endif %}