【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：

 1 /**

 2  * Definition for an interval.

 3  * public class Interval {

 4  *     int start;

 5  *     int end;

 6  *     Interval() { start = 0; end = 0; }

 7  *     Interval(int s, int e) { start = s; end = e; }

 8  * }

 9  */

10 public class Solution {

11     public List<Interval> insert(List<Interval> intervals, Interval newInterval) {

12         List<Interval> answer = new ArrayList<Interval>();

13         int insertPos = 0;

14         for(Interval in:intervals){

15             if(newInterval.start > in.end){

16                 answer.add(in);

17                 insertPos++;

18             }

19             else if(in.start > newInterval.end)

20             {

21                 answer.add(in);

22             }

23             else{

24                 newInterval.start = Math.min(in.start, newInterval.start);

25                 newInterval.end = Math.max(in.end, newInterval.end);

26             }            

27         }

28         answer.add(insertPos, newInterval);

29         return answer;

30     }

31 }