leetcode 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].

和上一题基本相同，不过本身是有序的，需要插入一个新的interval

思路是先把end<new.start的interval先放入，找到第一个end>=new.start的interval，然后在找到第一个start>new.end的interval，放入new，

再把剩余的interval逐个放入

/** * 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> res = new ArrayList<Interval>();        if(intervals.size()<1){            res.add(newInterval);            return res;        }        int i = 0;        while(i<intervals.size()&&intervals.get(i).end<newInterval.start){            res.add(intervals.get(i));            i++;        }        if(i==intervals.size()){            res.add(newInterval);            return res;        }        newInterval.start = Math.min(newInterval.start,intervals.get(i).start);        while(i<intervals.size() && intervals.get(i).start<=newInterval.end)          {              newInterval.end = Math.max(newInterval.end, intervals.get(i).end);              i++;          }        res.add(newInterval);        while(i<intervals.size())          {              res.add(intervals.get(i));              i++;          }      return res;    }}