June-20-2019

以前做的有问题,感觉遇到重复的会错,但是parameter变了,不好测了= =

这个题还是值得记一下的。其实是插入多个[1,1]这种interval。无非是跟前后判断就那么几种情况:

  • 左右相连,那一起MREGE,然后删掉后面的
  • 只和一边连,
  • 都不连
  • 其中一边囊括,

更重要等是一些general java 的相关

TreeMap

  • Key是排序的,可自定义comparator
  • lowerKey(), higherKey() 返还上一个,下一个存在的key
  • get(), remove(), containsKey(), lowerKey(), higherKey()都是O(lgN)

Stream

  • toArray(int[][]::new)

addNum: O(lgN) 做了好几次lgN: containsKey, lowerKey, higherKey, get()都是lgN

getInterval: O(N) 取决于最后有多少个interval

class SummaryRanges {
TreeMap<Integer, int[]> map;
/** Initialize your data structure here. */
public SummaryRanges() {
map = new TreeMap<>();
} public void addNum(int val) {
if (map.containsKey(val)) return;
Integer prevKey = map.lowerKey(val);
Integer nextKey = map.higherKey(val); if (prevKey != null && map.get(prevKey)[1] + 1 == val
&& nextKey != null && map.get(nextKey)[0] - 1 == val) {
map.get(prevKey)[1] = map.get(nextKey)[1];
map.remove(nextKey);
} else if (prevKey != null && map.get(prevKey)[1] + 1 == val) {
map.get(prevKey)[1] = val;
} else if (nextKey != null && map.get(nextKey)[0] - 1 == val) {
map.put(val, new int[]{val, map.get(nextKey)[1]});
map.remove(nextKey);
} else if (prevKey != null && map.get(prevKey)[1] >= val){
return;
} else {
map.put(val, new int[]{val, val});
} } public int[][] getIntervals() {
return map.entrySet()
.stream()
.map(entry -> new int[]{entry.getValue()[0], entry.getValue()[1]})
.toArray(int[][]::new); }
} /**
* Your SummaryRanges object will be instantiated and called as such:
* SummaryRanges obj = new SummaryRanges();
* obj.addNum(val);
* int[][] param_2 = obj.getIntervals();
*/

Follow up: What if there are lots of merges and the number of disjoint intervals are small compared to the data stream's size?

lots of merges说得很隐晦,其实指的是,很多次的进行add() -> 需要addNum()快一些。

log(n)已经很快了。。不知道怎么提速

然后最后interval少, getInterval()的 O(N)里N代表多少个区间,也很小,正好。。

(FB) 脸家给的follow-up是,addInterval(),不再是[2,2]这种,可以是[2,6]

楞做的话就 treeMap按照start排序

分情况

  • 完全包含在现有区间->无视
  • 完全不包含->加入新区间
  • 连接前后2个区间 -> merge
  • 和前后其中1个包含 -> 前后merge
  • 包含好几个现有区间 -> 最小start,最大end

仔细想想也挺麻烦。。

352. Data Stream as Disjoint Intervals的更多相关文章

  1. [LeetCode] 352. Data Stream as Disjoint Intervals 分离区间的数据流

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  2. leetcode@ [352] Data Stream as Disjoint Intervals (Binary Search & TreeSet)

    https://leetcode.com/problems/data-stream-as-disjoint-intervals/ Given a data stream input of non-ne ...

  3. 【leetcode】352. Data Stream as Disjoint Intervals

    问题描述: Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers ...

  4. 352. Data Stream as Disjoint Intervals (TreeMap, lambda, heapq)

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  5. [leetcode]352. Data Stream as Disjoint Intervals

    数据流合并成区间,每次新来一个数,表示成一个区间,然后在已经保存的区间中进行二分查找,最后结果有3种,插入头部,尾部,中间,插入头部,不管插入哪里,都判断一下左边和右边是否能和当前的数字接起来,我这样 ...

  6. 352[LeetCode] Data Stream as Disjoint Intervals

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  7. [LeetCode] Data Stream as Disjoint Intervals 分离区间的数据流

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  8. Leetcode: Data Stream as Disjoint Intervals && Summary of TreeMap

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  9. [Swift]LeetCode352. 将数据流变为多个不相交间隔 | Data Stream as Disjoint Intervals

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

随机推荐

  1. 从外国html5网站上扒来一个鼠标经过的css3 效果,感觉很不错

    鼠标经过的时候,感觉有点像一张纸卷上去的感觉. 下面是代码 <div class="main-container types"> <div class=" ...

  2. Java学习----日期函数

    Date   Calendar public class TestDate { private Date date; Calendar calendar; public TestDate() { // ...

  3. JavaScript 高阶函数 + generator生成器

    map/reduce map()方法定义在JavaScript的Array中,我们调用Array的map()方法,传入我们自己的函数,就得到了一个新的Array作为结果: function pow(x ...

  4. 将VIM配置成强大的IDE(三)

    上一节,我们知道了,我们了解了怎么配置插件的下下载. 现在,我们就可以去DIY我们的IDE了,主要介绍taglist插件和NERDTree插件,最终的结果是: 1.安装Taglist插件. Tagli ...

  5. Uninstall office15 click-to-run extensibility Component

    Summary : Uninstall office15 click-to-run extensibility Component,How to resolve Uninstall office15 ...

  6. Hadoop, Python, and NoSQL lead the pack for big data jobs

    Hadoop, Python, and NoSQL lead the pack for big data jobs   Rise in cloud-based analytics could incr ...

  7. iOS: 属性列表介绍 Introduction to Property Lists

    iOS: 属性列表介绍 Introduction to Property Lists 从本质上说, 属性列表就是苹果的对象数据序列化与反序列化方式 属性列表使用几种数据类型把数据组织为键值表和值表 P ...

  8. OOS升级服务

    给我们的应用程序做个版本更新服务,展示一个安装程序如何实现自动更新. //服务组,添加需要的任何服务 public enum ServerEnum { AutoupdateService,//自动升级 ...

  9. 抓取天涯文章的蜘蛛代码,刚经过更新(因为天涯页面HTML代码变化)

    #_*_coding:utf-8-*- import urllib2 import traceback import codecs from BeautifulSoup import Beautifu ...

  10. 十大响应式Web设计框架

    http://www.csdn.net/article/2014-05-13/2819739-responsive-frameworks-for-web-design 对于设计师而言,网站设计中的任意 ...