57. Insert Interval

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]

 

/**
 * 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> res = new ArrayList<Interval>();
        int len = intervals.size();
        int i=0;
        while(i<len&&intervals.get(i).end<newInterval.start){
            res.add(intervals.get(i++));
        }
        while(i<len&&intervals.get(i).start<=newInterval.end){
            newInterval.start = Math.min(newInterval.start,intervals.get(i).start);
            newInterval.end = Math.max(newInterval.end,intervals.get(i).end);
            i++;
        }
        res.add(newInterval);
        while(i<len){
            res.add(intervals.get(i++));
        }
        return res;
    }
}
//the run time complexity could be O(n),the space complexity could be O(n);