Leetcode_729. My Calendar I
阿新 • • 發佈:2018-12-24
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
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 toMyCalendar.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]
.
由於
start
and end
are
integers in the range [0, 10^9]
.
所以不能用陣列
可以用TreeMap 中的方法:
1 . floorKey()返回小於或者等於給定鍵的最大鍵,如果不存在返回null;
2 . ceilingKey() 返回大於等於給定鍵的最小值,如果不存在返回null;
思路: 1.用floorKey(start) 獲取 最靠近的start 左邊的key1 根據key1的value(end)大於start 則交叉 返回false
2. 用ceilingKey() 獲取
最靠近的start 右邊的key2 根據key 小於 start的end 則交叉 返回 false
class MyCalendar {
TreeMap<Integer, Integer> map;
public MyCalendar() {
map = new TreeMap<>();
}
public boolean book(int start, int end) {
Integer floorKey = map.floorKey(start);
if (floorKey != null && map.get(floorKey) > start) return false;
Integer ceilingKey = map.ceilingKey(start);
if (ceilingKey != null && ceilingKey < end) return false;
map.put(start, end);
return true;
}
}
參考: https://discuss.leetcode.com/topic/111205/java-8-liner-treemap