57. Insert Interval

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题目：
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].
思路：这题还是很简单的，根据例子就应该想到一种解决方案。
关键是找到newStart和newEnd。代码写的很烂

/** * 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> result = new ArrayList<>();        if(intervals.size() == 0)        {            result.add(newInterval);            return result;        }        int newStart = -1,newEnd = -1;        int flag = 0;        for(Interval interval : intervals)        {            int start = interval.start;            int end = interval.end;            if(newInterval.start>=start && newInterval.start<=end)            {                newStart = start;            }            if(newInterval.end>=start && newInterval.end<=end)            {                newEnd = end;            }        }        if(newStart == -1){newStart = newInterval.start;}        if(newEnd == -1){newEnd = newInterval.end;}        for(Interval interval : intervals)        {            if(interval.end<newStart)            {                result.add(interval);            }else if(flag == 0)            {                result.add(new Interval(newStart, newEnd));                flag = 1;            }            if(interval.start>newEnd)            {                result.add(interval);            }        }        if(flag == 0) result.add(newInterval);        return result;    }}

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