352. 将数据流变为多个不相交区间

给定一个非负整数的数据流输入 a1,a2,…,an,…,将到目前为止看到的数字总结为不相交的区间列表。

例如,假设数据流中的整数为 1,3,7,2,6,…,每次的总结为:

[1, 1]

[1, 1], [3, 3]

[1, 1], [3, 3], [7, 7]

[1, 3], [7, 7]

[1, 3], [6, 7]

进阶:

如果有很多合并,并且与数据流的大小相比,不相交区间的数量很小,该怎么办?

提示:

特别感谢 @yunhong 提供了本问题和其测试用例。

class SummaryRanges {

    private List<int[]> list = new ArrayList<>();

    /** Initialize your data structure here. */
public SummaryRanges() {
} public void addNum(int val) {
if (list.size() == 0) {
int[] arr = new int[2];
arr[0] = arr[1] = val;
list.add(arr);
return;
} int insertPosition = findInsertPosition(val);
if (insertPosition == list.size()) {
if (val == list.get(list.size() - 1)[1] + 1) {
list.get(list.size() - 1)[1] = list.get(list.size() - 1)[1] + 1;
} else {
int[] arr = new int[2];
arr[0] = arr[1] = val;
list.add(arr);
}
} else if (insertPosition == 0) {
if (val == list.get(0)[0] - 1) {
list.get(0)[0] = list.get(0)[0] - 1;
} else {
int[] arr = new int[2];
arr[0] = arr[1] = val;
list.add(0, arr);
}
} else if (insertPosition > 0) {
if (val == list.get(insertPosition)[0] - 1 && val == list.get(insertPosition - 1)[1] + 1) {
int index = insertPosition - 1;
int[] front = list.get(index);
int[] behind = list.get(insertPosition); list.remove(index);
list.remove(index); int[] arr = new int[2];
arr[0] = front[0];
arr[1] = behind[1];
list.add(index, arr);
} else if (val == list.get(insertPosition)[0] - 1) {
list.get(insertPosition)[0] = list.get(insertPosition)[0] - 1;
} else if (val == list.get(insertPosition - 1)[1] + 1) {
list.get(insertPosition - 1)[1] = list.get(insertPosition - 1)[1] + 1;
} else {
int[] arr = new int[2];
arr[0] = arr[1] = val;
list.add(insertPosition, arr);
}
}
} public int[][] getIntervals() {
int[][] result = new int[list.size()][]; for (int i = 0; i < list.size(); i++) {
result[i] = list.get(i);
}
return result;
} private int findInsertPosition(int val) {
int left = 0;
int right = list.size(); while (left < right) {
int mid = (left + right) / 2;
if (val >= list.get(mid)[0] && val <= list.get(mid)[1]) return -1; if (val < list.get(mid)[0]) right = mid;
else if (val > list.get(mid)[1]) left = mid + 1;
} return left;
}
} /**
* Your SummaryRanges object will be instantiated and called as such:
* SummaryRanges obj = new SummaryRanges();
* obj.addNum(val);
* int[][] param_2 = obj.getIntervals();
*/

Java实现 LeetCode 352 将数据流变为多个不相交区间的更多相关文章

  1. LeetCode352 将数据流变为多个不相交区间

    LeetCode352 将数据流变为多个不相交区间 1 题目 给你一个由非负整数 a1, a2, ..., an 组成的数据流输入,请你将到目前为止看到的数字总结为不相交的区间列表. 实现 Summa ...

  2. [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 ...

  3. [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 ...

  4. Java for LeetCode 023 Merge k Sorted Lists

    Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 解 ...

  5. Java for LeetCode 216 Combination Sum III

    Find all possible combinations of k numbers that add up to a number n, given that only numbers from ...

  6. Java for LeetCode 214 Shortest Palindrome

    Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. ...

  7. Java for LeetCode 212 Word Search II

    Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...

  8. Java for LeetCode 211 Add and Search Word - Data structure design

    Design a data structure that supports the following two operations: void addWord(word)bool search(wo ...

  9. Java for LeetCode 210 Course Schedule II

    There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prer ...

随机推荐

  1. 自动化测试po模式是什么?自动化测试po分层如何实现?-附详细源码

    一.什么是PO模式 全称:page object model  简称:POM/PO PO模式最核心的思想是分层,实现松耦合!实现脚本重复使用,实现脚本易维护性! ​ 主要分三层: 1.基础层BaseP ...

  2. [hdu3068 最长回文]Manacher算法,O(N)求最长回文子串

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3068 题意:求一个字符串的最长回文子串 思路: 枚举子串的两个端点,根据回文串的定义来判断其是否是回文 ...

  3. CODING 敏捷实战系列课第三讲:可视化业务分析

    业务分析处在开发过程的上游,提高业务分析的质量,可以减少后续开发.测试和集成过程中的反复确认,场景遗漏.采用可视化的业务分析工具箱可以大幅度避免文字版的业务需求描述所带来的不够完整,有误解等问题.CO ...

  4. 我参与 Seata 开源项目的一些感悟

    丁老师在他的知识星球邀请我回答以下一个问题: 我觉得这个问题非常有意思,姑且把它贴到公众号这里,与大家分享一下我对这个问题的一些感悟. 感谢丁老师的邀请问答: 在这里我就简单说下,我这段时间参与 Se ...

  5. 【雕爷学编程】Arduino动手做(61)---电压检测传感器

    37款传感器与执行器的提法,在网络上广泛流传,其实Arduino能够兼容的传感器模块肯定是不止这37种的.鉴于本人手头积累了一些传感器和执行器模块,依照实践出真知(一定要动手做)的理念,以学习和交流为 ...

  6. 最短Hamilton路径 数位dp

    最短Hamilton路径 #include<bits/stdc++.h> using namespace std; ; <<maxn][maxn]; int maps[maxn ...

  7. DRF节流组件

    1.DRF节流组件自定义(限制访问频率)  方式一 自定义类和方法: 和上述的认证组件使用方式一样,定义一个频率组件类,推荐继承BaseThrottle类, 需定义defallow_request(s ...

  8. 09 基于模块wsgiref版web框架

    09 基于模块wsgiref版web框架 模块引入 真实开发中的python web程序,一般会分为两部分:       服务器程序:负责对socket服务器进行封装,并在请求到来时,对请求的各种数据 ...

  9. PAT-1018 Public Bike Management(dijkstra + dfs)

    1018. Public Bike Management There is a public bike service in Hangzhou City which provides great co ...

  10. LightOJ1236

    题目大意: 给你一个 n,请你找出共有多少对(i,j)满足 lcm(i,j) = n (i<=j) . 解题思路: 我们利用算术基本定理将 n,i,j 进行分解: n = P1a1 * P2a2 ...