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;
}
}