[leetcode] 253. Meeting Rooms II 解题报告
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题目链接:https://leetcode.com/problems/meeting-rooms-ii/
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...]
(si < ei), find the minimum number of conference rooms required.
For example,
Given [[0, 30],[5, 10],[15, 20]]
,
return 2
.
思路:可以按照每个区间的起点排序, 然后依次将每个区间的终点放到一个集合中,如果下一个区间起点大于等于之前某区间的终点, 就将其从集合中删除. 这种做法的原理是如果一个区间的起点大于等于另一个区间的终点, 那么他们是不重合的.
代码如下:
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */class Solution {public: int minMeetingRooms(vector<Interval>& intervals) { auto cmp = [](Interval a, Interval b) { return a.start<b.start;}; sort(intervals.begin(), intervals.end(), cmp); multiset<int> st; int Max = 0; for(auto val: intervals) { while(!st.empty()&&val.start>= *st.begin()) st.erase(st.begin()); st.insert(val.end); Max = max(Max, (int)st.size()); } return Max; }};
还有另一个比较简洁的方法. 每个区间的起始点代表一个区间的开始,会将重叠区域+1,每个区间的结束点代表一个区间的结束,将会使重叠区域-1,因此我们可以利用这个性质,并结合STL的map来实现
代码如下:
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */class Solution {public: int minMeetingRooms(vector<Interval>& intervals) { map<int, int> hash; for(auto val: intervals) hash[val.start]++, hash[val.end]--; int ans = 0, sum = 0; for(auto val: hash) sum += val.second, ans = max(ans, sum); return ans; }};参考:https://leetcode.com/discuss/58720/c-solution-using-a-map-total-11-lines
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