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  • 【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:

     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 }
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  • 原文地址:https://www.cnblogs.com/sunshineatnoon/p/3868589.html
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