A Range Module is a module that tracks ranges of numbers. Your task is to design and implement the following interfaces in an efficient manner.

  • addRange(int left, int right) Adds the half-open interval [left, right), tracking every real number in that interval. Adding an interval that partially overlaps with currently tracked numbers should add any numbers in the interval [left, right) that are not already tracked.
  • queryRange(int left, int right) Returns true if and only if every real number in the interval [left, right) is currently being tracked.
  • removeRange(int left, int right) Stops tracking every real number currently being tracked in the interval [left, right).

Example 1:

addRange(10, 20): null
removeRange(14, 16): null
queryRange(10, 14): true (Every number in [10, 14) is being tracked)
queryRange(13, 15): false (Numbers like 14, 14.03, 14.17 in [13, 15) are not being tracked)
queryRange(16, 17): true (The number 16 in [16, 17) is still being tracked, despite the remove operation)

Note:

  • A half open interval [left, right) denotes all real numbers left <= x < right.
  • 0 < left < right < 10^9 in all calls to addRange, queryRange, removeRange.
  • The total number of calls to addRange in a single test case is at most 1000.
  • The total number of calls to queryRange in a single test case is at most 5000.
  • The total number of calls to removeRange in a single test case is at most 1000.
 

Approach #1: C++. [Using Vector]

class RangeModule {
public:
RangeModule() { } void addRange(int left, int right) {
vector<pair<int, int>> new_ranges;
bool inserted = false; for (const auto& it : ranges_) {
if (it.first > right && !inserted) {
new_ranges.emplace_back(left, right);
inserted = true;
}
if (it.first > right || it.second < left) {
new_ranges.push_back(it);
} else {
left = min(left, it.first);
right = max(right, it.second);
}
}
if (!inserted) new_ranges.emplace_back(left, right);
ranges_.swap(new_ranges);
} bool queryRange(int left, int right) {
int l = 0;
int r = ranges_.size() - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (ranges_[mid].first <= left && ranges_[mid].second >= right)
return true;
else if (ranges_[mid].first > right) {
r = mid - 1;
} else {
l = mid + 1;
}
}
return false;
} void removeRange(int left, int right) {
vector<pair<int, int>> new_ranges;
for (const auto& it : ranges_) {
if (it.second <= left || it.first >= right) {
new_ranges.emplace_back(it);
} else {
if (it.first < left)
new_ranges.emplace_back(it.first, left);
if (it.second > right)
new_ranges.emplace_back(right, it.second);
}
}
ranges_.swap(new_ranges);
} private:
vector<pair<int, int>> ranges_;
}; /**
* Your RangeModule object will be instantiated and called as such:
* RangeModule obj = new RangeModule();
* obj.addRange(left,right);
* bool param_2 = obj.queryRange(left,right);
* obj.removeRange(left,right);
*/

  

there are some notes about STL.

1. the difference between emplace_back and push_back.

2. emplace vs insert.

3. the method of swap in vector.

Approach #2: C++. [map]

class RangeModule {
public:
RangeModule() { } void addRange(int left, int right) {
IT l, r;
getOverLapRanges(left, right, l, r); if (l != r) {
auto last = r; last--;
left = min(left, l->first);
right = max(right, last->second);
ranges_.erase(l, r);
}
ranges_[left] = right;
} bool queryRange(int left, int right) {
IT l, r;
getOverLapRanges(left, right, l, r);
if (l == r) return false;
return l->first <= left && l->second >= right;
} void removeRange(int left, int right) {
IT l, r;
getOverLapRanges(left, right, l, r);
if (l == r) return ;
auto last = r; last--;
int start = min(left, l->first);
int end = max(right, last->second);
ranges_.erase(l, r);
if (start < left) ranges_[start] = left;
if (end > right) ranges_[right] = end;
} private:
typedef map<int, int>::iterator IT;
map<int, int> ranges_;
void getOverLapRanges(int left, int right, IT& l, IT& r) {
l = ranges_.upper_bound(left);
r = ranges_.upper_bound(right); // judge the left is the leftmost interval?
if (l != ranges_.begin()) {
if ((--l)->second < left) l++;
}
}
}; /**
* Your RangeModule object will be instantiated and called as such:
* RangeModule obj = new RangeModule();
* obj.addRange(left,right);
* bool param_2 = obj.queryRange(left,right);
* obj.removeRange(left,right);
*/

  

Notes:

c++::map.erase().

715. Range Module的更多相关文章

  1. Range Module

    2019-09-21 18:54:16 715. Range Module 问题描述: 问题求解: 用线段树解决了. class RangeModule { Node root; class Node ...

  2. [LeetCode] Range Module 范围模块

    A Range Module is a module that tracks ranges of numbers. Your task is to design and implement the f ...

  3. [Swift]LeetCode715. Range 模块 | Range Module

    A Range Module is a module that tracks ranges of numbers. Your task is to design and implement the f ...

  4. Java实现 LeetCode 715 Range 模块(选范围)

    715. Range 模块 Range 模块是跟踪数字范围的模块.你的任务是以一种有效的方式设计和实现以下接口. addRange(int left, int right) 添加半开区间 [left, ...

  5. 合并区间 · Merge Intervals & 插入区间 · Insert Interval

    [抄题]: 给出若干闭合区间,合并所有重叠的部分. 给出的区间列表 => 合并后的区间列表: [ [ [1, 3], [1, 6], [2, 6], => [8, 10], [8, 10] ...

  6. LeetCode All in One题解汇总(持续更新中...)

    突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对 ...

  7. leetcode 学习心得 (4)

    645. Set Mismatch The set S originally contains numbers from 1 to n. But unfortunately, due to the d ...

  8. All LeetCode Questions List 题目汇总

    All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems clas ...

  9. leetcode hard

    # Title Solution Acceptance Difficulty Frequency     4 Median of Two Sorted Arrays       27.2% Hard ...

随机推荐

  1. 我的Java开发学习之旅------>Java经典排序算法之插入排序

    一.算法原理 插入排序法:所谓插入排序法乃是将一个数目插入该占据的位置. 假设我们输入的是 "53,27,36,15,69,  42" 我们从第二个数字开始,这个数字是27,我们的 ...

  2. Java for LeetCode 090 Subsets II

    Given a collection of integers that might contain duplicates, nums, return all possible subsets. Not ...

  3. Flask中的模板语法jinjia2

    Flask中默认的模板语言是Jinja2 I. Jinja2模板语言中的 for {% for foo in g %} {% endfor %} II. Jinja2模板语言中的 if {% if g ...

  4. Spring事务超时时间可能存在的错误认识

    摘自:http://jinnianshilongnian.iteye.com/blog/1986023, 感谢作者. 1.先看代码 1.1.spring-config.xml <bean id= ...

  5. Spring Boot2.0之多数据源事务管理

    结合前面做的小项目,如果我把test01 test02下面的 service 都加了 事务的注解 这样启动时候会报错! 事务管理器里面不能有两个事务!!!! 这时候需要用  @Transactiona ...

  6. JSON.stringify出现 "Converting circular structure to JSON"

    JSON.stringify()  我们很熟悉了,将一个对象转换为json形式的字符串. 但是如果你在浏览器控制台中输出 JSON.stringify(window). 如果期望输出一段文字, 可能会 ...

  7. Android6.0 旋转屏幕(五)WMS启动应用流程(屏幕方向相关)

    一.强制设置方向 1.Activity 如果要强制设置一个Activity的横竖屏可以通过Manifest去设置,跟Activity相关的信息都会保存在ActivityInfo当中. android: ...

  8. BZOJ 1572 [Usaco2009 Open]工作安排Job:贪心 + 优先队列【先放再更新】

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1572 题意: 有n个工作,每个工作有一个截止日期dead[i]和收益pay[i]. 完成一 ...

  9. 图解mysql join

    原文:http://www.codeproject.com/Articles/33052/Visual-Representation-of-SQL-Joins 这个图文解释mysql join的各种技 ...

  10. kvm初体验之七:attach usb storage device to a VM

    1. virsh attach-disk vm1 /dev/sdb sdc 将host上的/dev/sdb挂载到vm1的/dev/sdc上 2. virsh detach-disk vm1 sdc 将 ...