57. Insert Interval

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].

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/** * 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> re = new ArrayList<>();int i = 0;for (; i < intervals.size(); ++i) {Interval temp = intervals.get(i);if (temp.end >= newInterval.start)break;elsere.add(temp);}Interval add = new Interval(newInterval.start, newInterval.end);for (; i < intervals.size(); ++i) {Interval temp = intervals.get(i);if (temp.end < newInterval.start || temp.start > newInterval.end)break;else {add.start = Math.min(add.start, temp.start);add.end = Math.max(add.end, temp.end);}}re.add(add);for (; i < intervals.size(); ++i)re.add(intervals.get(i));return re;    }}