- Time: 2015.01.17
- Tags: simulation
Problem Description link
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 Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
List<Interval> set = new LinkedList<Interval>();
if (newInterval == null){
return intervals;
}
boolean added = false;
for (Interval v : intervals){
if (v.end < newInterval.start){
set.add(v);
}else if (v.start > newInterval.end){
if (!added){
set.add(newInterval);
added = true;
}
set.add(v);
}else{
newInterval.start = Math.min(newInterval.start, v.start);
newInterval.end = Math.max(newInterval.end, v.end);
}
}
if (!added){
set.add(newInterval);
}
return set;
}
}