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You are given an integer n
. There are n
rooms numbered from 0
to n - 1
.
You are given a 2D integer array meetings
where meetings[i] = [starti, endi]
means that a meeting will be held during the half-closed time interval [starti, endi)
. All the values of starti
are unique.
Meetings are allocated to rooms in the following manner:
Return the number of the room that held the most meetings. If there are multiple rooms, return the room with the lowest number.
A half-closed interval [a, b)
is the interval between a
and b
including a
and not including b
.
Example 1:
Input: n = 2, meetings = [[0,10],[1,5],[2,7],[3,4]] Output: 0 Explanation: - At time 0, both rooms are not being used. The first meeting starts in room 0. - At time 1, only room 1 is not being used. The second meeting starts in room 1. - At time 2, both rooms are being used. The third meeting is delayed. - At time 3, both rooms are being used. The fourth meeting is delayed. - At time 5, the meeting in room 1 finishes. The third meeting starts in room 1 for the time period [5,10). - At time 10, the meetings in both rooms finish. The fourth meeting starts in room 0 for the time period [10,11). Both rooms 0 and 1 held 2 meetings, so we return 0.
Example 2:
Input: n = 3, meetings = [[1,20],[2,10],[3,5],[4,9],[6,8]] Output: 1 Explanation: - At time 1, all three rooms are not being used. The first meeting starts in room 0. - At time 2, rooms 1 and 2 are not being used. The second meeting starts in room 1. - At time 3, only room 2 is not being used. The third meeting starts in room 2. - At time 4, all three rooms are being used. The fourth meeting is delayed. - At time 5, the meeting in room 2 finishes. The fourth meeting starts in room 2 for the time period [5,10). - At time 6, all three rooms are being used. The fifth meeting is delayed. - At time 10, the meetings in rooms 1 and 2 finish. The fifth meeting starts in room 1 for the time period [10,12). Room 0 held 1 meeting while rooms 1 and 2 each held 2 meetings, so we return 1.
Constraints:
1 <= n <= 100
1 <= meetings.length <= 105
meetings[i].length == 2
0 <= starti < endi <= 5 * 105
starti
are unique.We define two priority queues, representing idle meeting rooms and busy meeting rooms, respectively. Among them: the idle meeting rooms idle
are sorted according to index; while the busy meeting rooms busy
are sorted according to end time, index.
First, sort the meetings by start time, then traverse the meetings. For each meeting:
idle
.idle
and add it to the busy queue busy
.busy
, and re-add it to the busy queue busy
.The time complexity is $O(m \times \log m)$, where $m$ is the number of meetings.
Similar problems:
class Solution:
def mostBooked(self, n: int, meetings: List[List[int]]) -> int:
meetings.sort()
busy = []
idle = list(range(n))
heapify(idle)
cnt = [0] * n
for s, e in meetings:
while busy and busy[0][0] <= s:
heappush(idle, heappop(busy)[1])
if idle:
i = heappop(idle)
cnt[i] += 1
heappush(busy, (e, i))
else:
a, i = heappop(busy)
cnt[i] += 1
heappush(busy, (a + e - s, i))
ans = 0
for i, v in enumerate(cnt):
if cnt[ans] < v:
ans = i
return ans
class Solution {
public int mostBooked(int n, int[][] meetings) {
Arrays.sort(meetings, (a, b) -> a[0] - b[0]);
PriorityQueue<int[]> busy
= new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
PriorityQueue<Integer> idle = new PriorityQueue<>();
for (int i = 0; i < n; ++i) {
idle.offer(i);
}
int[] cnt = new int[n];
for (var v : meetings) {
int s = v[0], e = v[1];
while (!busy.isEmpty() && busy.peek()[0] <= s) {
idle.offer(busy.poll()[1]);
}
int i = 0;
if (!idle.isEmpty()) {
i = idle.poll();
busy.offer(new int[] {e, i});
} else {
var x = busy.poll();
i = x[1];
busy.offer(new int[] {x[0] + e - s, i});
}
++cnt[i];
}
int ans = 0;
for (int i = 0; i < n; ++i) {
if (cnt[ans] < cnt[i]) {
ans = i;
}
}
return ans;
}
}
using ll = long long;
using pii = pair<ll, int>;
class Solution {
public:
int mostBooked(int n, vector<vector<int>>& meetings) {
priority_queue<int, vector<int>, greater<int>> idle;
priority_queue<pii, vector<pii>, greater<pii>> busy;
for (int i = 0; i < n; ++i) idle.push(i);
vector<int> cnt(n);
sort(meetings.begin(), meetings.end());
for (auto& v : meetings) {
int s = v[0], e = v[1];
while (!busy.empty() && busy.top().first <= s) {
idle.push(busy.top().second);
busy.pop();
}
int i = 0;
if (!idle.empty()) {
i = idle.top();
idle.pop();
busy.push({e, i});
} else {
auto x = busy.top();
busy.pop();
i = x.second;
busy.push({x.first + e - s, i});
}
++cnt[i];
}
int ans = 0;
for (int i = 0; i < n; ++i) {
if (cnt[ans] < cnt[i]) {
ans = i;
}
}
return ans;
}
};
func mostBooked(n int, meetings [][]int) int {
sort.Slice(meetings, func(i, j int) bool { return meetings[i][0] < meetings[j][0] })
idle := hp{make([]int, n)}
for i := 0; i < n; i++ {
idle.IntSlice[i] = i
}
busy := hp2{}
cnt := make([]int, n)
for _, v := range meetings {
s, e := v[0], v[1]
for len(busy) > 0 && busy[0].end <= s {
heap.Push(&idle, heap.Pop(&busy).(pair).i)
}
var i int
if idle.Len() > 0 {
i = heap.Pop(&idle).(int)
heap.Push(&busy, pair{e, i})
} else {
x := heap.Pop(&busy).(pair)
i = x.i
heap.Push(&busy, pair{x.end + e - s, i})
}
cnt[i]++
}
ans := 0
for i, v := range cnt {
if cnt[ans] < v {
ans = i
}
}
return ans
}
type hp struct{ sort.IntSlice }
func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}
type pair struct{ end, i int }
type hp2 []pair
func (h hp2) Len() int { return len(h) }
func (h hp2) Less(i, j int) bool {
a, b := h[i], h[j]
return a.end < b.end || a.end == b.end && a.i < b.i
}
func (h hp2) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp2) Push(v any) { *h = append(*h, v.(pair)) }
func (h *hp2) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
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