LeetCode 729. My Calendar I 红黑树 线段树 二分查找树

Implement a MyCalendar class to store your events. A new event can be added if adding the event will not cause a double booking.

Your class will have the method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.

A double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false and do not add the event to the calendar.

Your class will be called like this: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)

Example 1:

MyCalendar();
MyCalendar.book(10, 20); // returns true
MyCalendar.book(15, 25); // returns false
MyCalendar.book(20, 30); // returns true
Explanation: 
The first event can be booked.  The second can't because time 15 is already booked by another event.
The third event can be booked, as the first event takes every time less than 20, but not including 20.

 

Note:

• The number of calls to MyCalendar.book per test case will be at most 1000.
• In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

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方案一：c++ 红黑树

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
using namespace std;

class MyCalendar {
    map<int, int> books;
public:
    bool book(int s, int e) {
        auto next = books.lower_bound(s); // first element with key not go before k (i.e., either it is equivalent or goes after).
        if (next != books.end() && next->first < e) return false; // a existing book started within the new book (next)
        if (next != books.begin() && s < (--next)->second) return false; // new book started within a existing book (prev)
        books[s] = e;
        return true;
    }
};

int main()
{
    MyCalendar my;
    bool r1 = my.book(10, 20); // returns true
    bool r2 = my.book(15, 25); // returns false
    bool r3 = my.book(20, 30); // returns true
    bool r4 = my.book(12, 13);
    return 0;
}

方案二：python 手写BST

class Node:
    def __init__(self, s, e):
        self.s, self.e = s, e
        self.l, self.r = None, None


class MyCalendar:

    def __init__(self):
        self.rt = None

    def book(self, start: int, end: int) -> bool:
        def dfs(s, e, node):
            if (s >= node.e):
                if (node.r):
                    return dfs(s, e, node.r)
                else:
                    node.r = Node(s, e)
                    return True
            elif (e <= node.s):
                if (node.l):
                    return dfs(s, e, node.l)
                else:
                    node.l = Node(s, e)
                    return True
            else:
                return False

        if (self.rt == None):
            self.rt = Node(start, end)
            return True
        else:
            return dfs(start, end, self.rt)