LWC 59：729. My Calendar I

LWC 59：729. My Calendar I

传送门：729. My Calendar I

Problem:

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].

思路： 判断新加入的区间是否与原来的重叠，总共有三种重叠情况，左相交，右相交和包含，如下图：

alt text

当然这里还有一种情况，即黄色块完全包含于红色块，但这种情况已经包含在了A情况或者B情况中，所以不需要重复判断。

代码如下：

class MyCalendar {

    class Interval{
        int s;
        int e;

        Interval(int s, int e){
            this.s = s;
            this.e = e;
        }
    }

    List<Interval> mem;

    public MyCalendar() {
        mem = new ArrayList<>();
    }

    public boolean book(int start, int end) {
        Interval candicate = new Interval(start, end);
        if (overlap(candicate)) {
            return false;
        }
        else {
            mem.add(candicate);
            return true;
        }
    }

    public boolean overlap(Interval cand) {
        for (Interval tmp : mem) {
            if (overlap(tmp, cand)) return true;
        }
        return false;
    }

    boolean overlap(Interval a, Interval b) {
        if (b.s >= a.s && b.s < a.e || b.e <= a.e && b.e > a.s || b.s <= a.s && b.e >= a.e) {
            return true;
        }
        return false;
    }
}