Description

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].

Solution

Iteration, time O(n), space O(n)

本来想用二分查找做，感觉有点麻烦，后来发现还不如一个一个扫过去简单。创建一个List<Interval> res来存储merge后的intervals比较简单，不用再原链表上做修改。

/**
 * 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 (isOverlap(newInterval, interval)) {
                newInterval = merge(newInterval, interval);
            } else {
                if (interval.end < newInterval.start) {
                    res.add(interval);
                } else {
                    res.add(newInterval);
                    newInterval = interval;
                }
            }
        }
        
        res.add(newInterval);
        return res;
    }
    
    public boolean isOverlap(Interval a, Interval b) {
        if (a.start > b.start) {
            return isOverlap(b, a);
        }
        return a.end >= b.start;
    }
    
    public Interval merge(Interval a, Interval b) {
        return new Interval(Math.min(a.start, b.start), Math.max(a.end, b.end));
    }
}