我的日程表 I(729. My Calendar I)
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2024-02-24 22:29:40
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1 描述
Description
- 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
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].
2 代码
#include <iostream>
#include <map>
using namespace std;
class MyCalendar {
public:
map<int, int> books;
MyCalendar() {}
bool book(int start, int end) {
bool res = false;
auto itS = books.end();
auto itE = books.end();
auto temp = books.begin();
while (temp != books.end())
{
if (start < temp->second)
{
itS = temp;
break;
}
temp++;
}
temp = books.end();
while (temp != itS)
{
--temp;
if (end > temp->first)
{
itE = temp;
break;
}
}
if (itE == books.end())
{
res = true;
books.insert(make_pair(start, end));
}
return res;
}
};
int main()
{
MyCalendar* obj = new MyCalendar();
bool res = obj->book(10, 20);
cout << res << endl;
res = obj->book(15, 25);
cout << res << endl;
res = obj->book(20, 30);
cout << res << endl;
res = obj->book(25, 35);
cout << res << endl;
system("pause");
return 0;
}
3 解释
主要思路,找出所有与 “新计划” 相冲突的 “旧计划”
- “旧计划” 个数为 0,则证明不冲突;
- “旧计划” 个数大于 0,则证明冲突;
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