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].
题意及分析：给出一个按升序排序的非重叠区间序列，插入一个新区间，合并有重叠的区间。
代码：
/** * 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; } * } */ class Solution { //给出一个按升序排序的非重叠区间序列，插入一个新区间，合并有重叠的区间 public List<Interval> insert(List<Interval> intervals, Interval newInterval) { List<Interval> res = new ArrayList<>(); int i=0; while(i<intervals.size() && intervals.get(i).end<newInterval.start){ //交叉区间之前的直接加入 res.add(intervals.get(i++)); } while(i<intervals.size() && intervals.get(i).start <= newInterval.end){ //有交叉区间 newInterval = new Interval( Math.min(intervals.get(i).start,newInterval.start), Math.max(intervals.get(i).end,newInterval.end) ); i++; } res.add(newInterval); while (i < intervals.size()) //重叠区间之后的也直接加入 res.add(intervals.get(i++)); return res; } }
