【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].

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;

     }

 }

巴特西

【leetcode刷题笔记】Insert Interval

最新文章

热门文章