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InsertInterval.java
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73 lines (64 loc) · 1.81 KB
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package algorithm.lc;
import java.util.ArrayList;
import java.util.Comparator;
/**
* 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]. *
*/
public class InsertInterval {
public class Interval {
int start;
int end;
Interval() {
start = 0;
end = 0;
}
Interval(int s, int e) {
start = s;
end = e;
}
}
// only one scan
public class Solution {
public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
// Start typing your Java solution below
// DO NOT write main() function
ArrayList<Interval> res = new ArrayList<Interval>();
int start = newInterval.start;
int end = newInterval.end;
boolean finished = false;
for (int i = 0; i < intervals.size(); ++i) {
Interval interval = intervals.get(i);
if (interval.end < start) {
res.add(interval);
}
else if (interval.start > end) {
if (finished == false) { // new interval has not been added
res.add(new Interval(start, end));
finished = true;
}
res.add(interval);
}
else {
start = Math.min(interval.start, start);
end = Math.max(interval.end, end);
}
}
if (finished == false) {
res.add(new Interval(start, end));
}
return res;
}
}
}