57 Insert Interval

57. Insert Interval

1. Question

Given a set ofnon-overlappingintervals, 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].

2. Implementation

(1) Scan Line

/**
 * 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; }
 * }
 */
class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> res = new ArrayList<>();

        for (Interval interval : intervals) {
            if (newInterval == null || interval.end < newInterval.start) {
                res.add(interval);
            }
            else if (newInterval.end < interval.start) {
                res.add(newInterval);
                res.add(interval);
                newInterval = null;
            }
            else {
                newInterval.start = Math.min(interval.start, newInterval.start);
                newInterval.end = Math.max(interval.end, newInterval.end);
            }
        }

        if (newInterval != null) {
            res.add(newInterval);
        }
        return res;
    }
}

3. Time & Space Complexity

Scan Line: 时间复杂度O(n), 空间复杂度O(n)