zoukankan      html  css  js  c++  java
  • [LintCode] Insert Interval

    Given a non-overlapping interval list which is sorted by start point.

    Insert a new interval into it, make sure the list is still in order and non-overlapping (merge intervals if necessary).

    Example

    Insert [2, 5] into [[1,2], [5,9]], we get [[1,9]].

    Insert [3, 4] into [[1,2], [5,9]], we get [[1,2], [3,4], [5,9]].

    Algorithm: 

    Iterate over the original interval list one by one and keep an insertion position for the new interval.

    Case 1: newInterval is after the current interval:  If the current interval's end < newInterval's start, insert the current interval to the new list, increment the insertion position by 1.

    Case 2: newInterval is before the current interval: Else if the current interval's start > newInterval's end, insert the current interval to the new list. 

    Case 3: newInterval overlaps with the current interval. Update the newInterval's start and end to cover both intervals. 

    Repeat the above steps until all intervals in the original list are processed. 

    Finally, insert the newInterval to the saved insertion position. 

     1 /**
     2  * Definition of Interval:
     3  * public classs Interval {
     4  *     int start, end;
     5  *     Interval(int start, int end) {
     6  *         this.start = start;
     7  *         this.end = end;
     8  *     }
     9  */
    10 public class Solution {
    11     public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
    12         if (newInterval == null) {
    13             return intervals;
    14         }
    15         ArrayList<Interval> results = new ArrayList<Interval>();
    16         if(intervals == null) {
    17             results.add(newInterval);
    18             return results;
    19         }
    20         int insertPos = 0;
    21         for (Interval interval : intervals) {
    22             if (interval.end < newInterval.start) {
    23                 results.add(interval);
    24                 insertPos++;
    25             } else if (interval.start > newInterval.end) {
    26                 results.add(interval);
    27             } else {
    28                 newInterval.start = Math.min(interval.start, newInterval.start);
    29                 newInterval.end = Math.max(interval.end, newInterval.end);
    30             }
    31         }
    32         results.add(insertPos, newInterval);
    33         return results;
    34     }
    35 }

    Related Problems 

    Merge Intervals

  • 相关阅读:
    搜索文件/目录的shell脚本
    git的编译安装
    linux命令行直接执行MySQL/MariaDB语句查询
    MySQL重置root密码
    图解TCP/IP三次握手
    使用pull命令从Docker Hub仓库中下载镜像到本地
    BZOJ1051 [HAOI2006]受欢迎的牛(Tarjan缩点)
    BZOJ1026 [SCOI2009]windy数(数位DP)
    CERC2017 H Hidden Hierarchy(树+模拟)
    2018icpc徐州网络赛-H Ryuji doesn't want to study(线段树)
  • 原文地址:https://www.cnblogs.com/lz87/p/7169574.html
Copyright © 2011-2022 走看看