【leetcode刷题笔记】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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题解：利用变量insertPos记录最终区间插入的位置。遍历intervals，那么当前遍历的interval和newInterval有以下几种情况：

代码如下：

 /**
  * 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> answer = new ArrayList<Interval>();
         int insertPos = 0;
         for(Interval in:intervals){
             if(newInterval.start > in.end){
                 answer.add(in);
                 insertPos++;
             }
             else if(in.start > newInterval.end)
             {
                 answer.add(in);
             }
             else{
                 newInterval.start = Math.min(in.start, newInterval.start);
                 newInterval.end = Math.max(in.end, newInterval.end);
             }
         }
         answer.add(insertPos, newInterval);
         return answer;
     }
 }