题目描述:
Implement a MyCalendarTwo
class to store your events. A new event can be added if adding the event will not cause a triple booking.
Your class will have one 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 triple booking happens when three events have some non-empty intersection (ie., there is some time that is common to all 3 events.)
For each call to the method MyCalendar.book
, return 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.
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(50, 60); // returns true MyCalendar.book(10, 40); // returns true MyCalendar.book(5, 15); // returns false MyCalendar.book(5, 10); // returns true MyCalendar.book(25, 55); // returns true Explanation: The first two events can be booked. The third event can be double booked. The fourth event (5, 15) can't be booked, because it would result in a triple booking. The fifth event (5, 10) can be booked, as it does not use time 10 which is already double booked. The sixth event (25, 55) 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.
Note:
- The number of calls to
MyCalendar.book
per test case will be at most1000
. - In calls to
MyCalendar.book(start, end)
,start
andend
are integers in the range[0, 10^9]
.
题目大意:
实现MyCalendarTwo类,存储一组活动。
函数book(start, end)输入一个活动的起止时间,若该活动至多与两个现有活动冲突,返回true。否则返回false。
解题思路:
利用HashMap存储各区间端点被覆盖的次数,记为dmap
利用List存储所有不triple booking的活动,记为events
每次添加活动时,遍历events,判断当前活动与已存在活动是否发生冲突,并更新dmap
将终点坐标减去0.5,与起点坐标做区分
Java代码:
class MyCalendarTwo {
private Map<Double, Integer> dmap;
private List<int[]> events;
public MyCalendarTwo() {
dmap = new HashMap<>();
events = new ArrayList<>();
}
public boolean book(int start, int end) {
int cs = 1, ce = 1;
for (int[] e: events) {
if (e[0] <= start && start < e[1]) cs++;
if (e[0] < end && end < e[1]) ce++;
}
if (cs > 2 || ce > 2) return false;
List<Double> addList = new ArrayList<>();
for (double key : dmap.keySet()) {
if (start <= key && key <= end - 0.5) {
if (dmap.get(key) == 2) return false;
addList.add(key);
}
}
for (Double key: addList) dmap.put(key, dmap.get(key) + 1);
if (!dmap.containsKey(start)) {
dmap.put((double)start, cs);
}
if (!dmap.containsKey(end - 0.5)) {
dmap.put(end - 0.5, ce);
}
events.add(new int[]{start, end});
return true;
}
}
/**
* Your MyCalendarTwo object will be instantiated and called as such:
* MyCalendarTwo obj = new MyCalendarTwo();
* boolean param_1 = obj.book(start,end);
*/
本文链接:http://bookshadow.com/weblog/2017/11/20/leetcode-my-calendar-ii/
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