Insert Interval - LeetCode

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].

这道题的作法就是，如果插入的interval和我们正在访问的interval无关，就把那个interval放到结果集里，如果有关，就对我们手上需要插入的interval进行更新。很多时候我们做一个新的空集放结果，比我们在原集上更改要方便的多（虽然浪费了一定的空间）。

AC代码如下：

public class Solution {    public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {         ArrayList<Interval> result = new ArrayList<Interval>();         for(Interval interval: intervals){            if(interval.end < newInterval.start){                result.add(interval);            }else if(interval.start > newInterval.end){                result.add(newInterval);                newInterval = interval;                    }else if(interval.end >= newInterval.start || interval.start <= newInterval.end){                newInterval = new Interval(Math.min(interval.start, newInterval.start), Math.max(newInterval.end, interval.end));            }        }         result.add(newInterval); //什么时候都不要忘记处理最后一个         return result;    }}