731. 我的日程安排表 II

🟠   🔖  设计 线段树 二分查找 有序集合  🔗 LeetCode

题目

You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a triple booking.

A triple booking happens when three events have some non-empty intersection (i.e., some moment is common to all the three events.).

The event can be represented as a pair of integers start and end that represents a booking on the half-open interval [start, end), the range of real numbers x such that start <= x < end.

Implement the MyCalendarTwo class:

• MyCalendarTwo() Initializes the calendar object.
• boolean book(int start, int end) Returns true if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return false and do not add the event to the calendar.

Example 1:

Input

["MyCalendarTwo", "book", "book", "book", "book", "book", "book"]

[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]

Output

[null, true, true, true, false, true, true]

Explanation

MyCalendarTwo myCalendarTwo = new MyCalendarTwo();

myCalendarTwo.book(10, 20); // return True, The event can be booked.

myCalendarTwo.book(50, 60); // return True, The event can be booked.

myCalendarTwo.book(10, 40); // return True, The event can be double booked.

myCalendarTwo.book(5, 15); // return False, The event cannot be booked, because it would result in a triple booking.

myCalendarTwo.book(5, 10); // return True, The event can be booked, as it does not use time 10 which is already double booked.

myCalendarTwo.book(25, 55); // return True, The event can be booked, as the time in [25, 40) will be double booked with the third event, the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

Constraints:

• 0 <= start < end <= 10^9
• At most 1000 calls will be made to book.

题目大意

实现一个 MyCalendar 类来存放你的日程安排。如果要添加的时间内不会导致三重预订时，则可以存储这个新的日程安排。

MyCalendar 有一个 book(int start, int end)方法。它意味着在 start 到 end 时间内增加一个日程安排，注意，这里的时间是半开区间，即 [start, end), 实数 x 的范围为， start <= x < end。

当三个日程安排有一些时间上的交叉时（例如三个日程安排都在同一时间内），就会产生三重预订。

每次调用 MyCalendar.book 方法时，如果可以将日程安排成功添加到日历中而不会导致三重预订，返回 true。否则，返回 false 并且不要将该日程安排添加到日历中。

请按照以下步骤调用 MyCalendar 类:

MyCalendar cal = new MyCalendar();
MyCalendar.book(start, end)

解题思路

1. 初始化一个空的 deltas 对象，用于记录时间点的变化。
2. book 方法：
   • 对于新事件 [start, end)，更新 deltas 对象，将 deltas[start] 加一， deltas[start] 减一。 - 按照时间点顺序遍历 deltas，计算当前重叠事件的数量。
   • 遍历所有事件，累加计算重叠数量，如果发现重叠数量达到 3，表示会导致三重预订，此时撤销之前对 deltas 的修改并返回 false。
   • 如果没有冲突，表示成功添加事件，返回 true。

复杂度分析

• 时间复杂度：O(n log n)，主要是因为我们需要对 deltas 的键进行排序。
• 空间复杂度：O(n)，用于存储 deltas 对象。

代码

var MyCalendarTwo = function () {
	this.deltas = {};
};

/**
 * @param {number} start
 * @param {number} end
 * @return {boolean}
 */
MyCalendarTwo.prototype.book = function (start, end) {
	// 更新重叠事件计数
	this.deltas[start] = (this.deltas[start] || 0) + 1; // 新事件开始
	this.deltas[end] = (this.deltas[end] || 0) - 1; // 新事件结束

	// 检查重叠数量
	let overlap = 0;
	for (const time of Object.keys(this.deltas).sort((a, b) => a - b)) {
		overlap += this.deltas[time];
		if (overlap >= 3) {
			// 如果重叠事件达到3，撤销之前的修改
			this.deltas[start] -= 1;
			this.deltas[end] += 1;
			if (this.deltas[start] === 0) delete this.deltas[start];
			if (this.deltas[end] === 0) delete this.deltas[end];
			return false; // 返回 false，表示添加失败
		}
	}
	return true; // 成功添加事件，返回 true
};

/**
 * Your MyCalendarTwo object will be instantiated and called as such:
 * var obj = new MyCalendarTwo()
 * var param_1 = obj.book(start,end)
 */

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