57 Insert Interval
Problem:
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].
Solutions:
/**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
public class Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
List<Interval> result = new LinkedList<Interval>();
boolean inserted = false;
for (Interval inter:intervals) {
if (inserted) {
result.add(inter);
}
else if (newInterval.start > inter.end) {
//no overlap
result.add(inter);
}
else if (newInterval.end < inter.start) {
inserted = true;
result.add(newInterval);
result.add(inter);
}
else {
newInterval.start = Math.min(newInterval.start, inter.start);
newInterval.end = Math.max(newInterval.end, inter.end);
}
}
if (!inserted) {
result.add(newInterval);
}
return result;
}
}