LeetCode 729/731. My Calendar I + II

My Calendar I

hash table + binary search，注意边界情况即可。

class MyCalendar {
private:
    map<int,int> m; // map[start] -> end
    
public:
    MyCalendar() {}
    
    bool book(int start, int end) {
        auto it=m.lower_bound(start);
        if (it!=m.end() && end>it->first) return false;
        if (it!=m.begin() && (--it)->second>start) return false;
        m[start] = end;
        return true;
    }
};

/**
 * Your MyCalendar object will be instantiated and called as such:
 * MyCalendar* obj = new MyCalendar();
 * bool param_1 = obj->book(start,end);
 */

 

My Calendar II

方法一：沿用 My Calendar I 的思路，给overlap的时间段再建一个hashtable，判断overlap的时间有没有overlap。

https://leetcode.com/problems/my-calendar-ii/discuss/109519/JavaC++-Clean-Code-with-Explanation

 

方法二：Boundary Count

思路与 Number of Meeting Rooms at a Given Time 一致，只不过每次book都要计算一遍所有时刻有没有符合要求。

class MyCalendarTwo {
private:
    map<int,int> m; // time->num
public:
    MyCalendarTwo() {}
    
    bool book(int start, int end) {
        ++m[start]; --m[end];
        int sum=0;
        for (auto it=m.begin();it!=m.end();++it){
            if (it->second+sum >= 3){
                if (--m[start]==0) m.erase(start); 
                ++m[end];
                return false;
            }
            sum += it->second;
        }
        return true;
    }
};

/**
 * Your MyCalendarTwo object will be instantiated and called as such:
 * MyCalendarTwo* obj = new MyCalendarTwo();
 * bool param_1 = obj->book(start,end);
 */

时间复杂度 O(n^2)