Leetcode 253. 會議室 II C++解法
阿新 • • 發佈:2021-01-25
技術標籤:LeetCode&CodeWarc++leetcode
Leetcode 253. 會議室 II
https://leetcode-cn.com/problems/meeting-rooms-ii/
給你一個會議時間安排的陣列 intervals ,每個會議時間都會包括開始和結束的時間 intervals[i] = [starti, endi] ,為避免會議衝突,同時要考慮充分利用會議室資源,請你計算至少需要多少間會議室,才能滿足這些會議安排。
示例 1:
輸入:intervals = [[0,30],[5,10],[15,20]]
輸出:2
示例 2:
輸入:intervals = [[7,10],[2,4]] 輸出:1
提示:
1 <= intervals.length <= 104
0 <= starti < endi <= 106
解法1. 使用優先佇列
class Solution {
public:
int minMeetingRooms(vector<vector<int>>& intervals) {
if (intervals.size() == 0) {
return 0;
}
priority_queue<int, std::vector<int >, std::greater<int> > allocator;
sort(intervals.begin(), intervals.end(),
[](auto a, auto b) { return a[0] < b[0]; });
allocator.push(intervals[0][1]);
for (int i = 1; i < intervals.size(); i++) {
if (intervals[i][0] >= allocator. top()) {
allocator.pop();
}
allocator.push(intervals[i][1]);
}
return allocator.size();
}
};
解法2. 有序化處理
class Solution {
public:
int minMeetingRooms(vector<vector<int>>& intervals) {
if (intervals.size() == 0) {
return 0;
}
vector<int> starts(intervals.size(), 0);
vector<int> ends(intervals.size(), 0);
for (int i = 0; i < intervals.size(); i++) {
starts[i] = intervals[i][0];
ends[i] = intervals[i][1];
}
sort(starts.begin(), starts.end());
sort(ends.begin(), ends.end());
auto startIter = starts.begin();
auto endIter = ends.begin();
int useRooms = 0 ;
while (startIter != starts.end()) {
if (*startIter >= *endIter) {
useRooms -= 1;
endIter++;
}
useRooms += 1;
startIter++;
}
return useRooms;
}
};