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;
    }
}