Insert Interval

题目描述：
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].

这个题不难，关键是要考虑周到，每一种情况都要想清楚。leetcode上对于不会通过的case都会有错误提示，但是很多公司招聘的oj平台是不会给错误提示的，所以大家要试着不去看错误提示，加油吧。

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {    List<Interval> list=new ArrayList<Interval>();    int n=intervals.size();    if(n==0){        list.add(newInterval);    }    boolean go=false;    for (int i = 0; i < n; i++) {        if(go){            list.add(intervals.get(i));break;        }        if(intervals.get(i).end<newInterval.start){            list.add(intervals.get(i));            //当newInterval是最后一个的时候，要直接插入            if(i==n-1){                list.add(newInterval);            }                              }else if(intervals.get(i).start>newInterval.end){            list.add(intervals.get(i));            list.add(newInterval);            go=true;        }else{            newInterval.start=newInterval.start<intervals.get(i).start?newInterval.start:intervals.get(i).start;            newInterval.end=newInterval.end>intervals.get(i).end?newInterval.end:intervals.get(i).end;            //当newInterval是最后一个的时候，要直接插入            if(i==n-1){                list.add(newInterval);            }        }    }    return list;}