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].

题目中给定了一个排好序的interval数组，interval之间没有覆盖，然后插入一个新的interval，merge之后返回，从第一个元素开始，如果有冲突更新newInterval的start，将它加入结果集中，然后判断newInterval的end属性。遍历一遍数组，时间复杂度为O(n)。代码如下：

/**
 * 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> list = new ArrayList<Interval>();
        if(intervals == null || intervals.size() == 0) {
            list.add(newInterval);
            return list;
        }
        int i = 0;
        while(i < intervals.size() && intervals.get(i).end < newInterval.start) {
            list.add(intervals.get(i));
            i ++;
        }
        if(i < intervals.size()) 
            newInterval.start = Math.min(newInterval.start, intervals.get(i).start);
        list.add(newInterval);
        while(i < intervals.size() && newInterval.end >= intervals.get(i).start) {
            newInterval.end = Math.max(newInterval.end, intervals.get(i).end);
            i ++;
        }
        while (i < intervals.size()) {
            list.add(intervals.get(i));
            i ++;
        }
        return list;
    }
}