README_EN.md

253. Meeting Rooms II

中文文档

Description

Given an array of meeting time intervals intervals where intervals[i] = [starti, endi], return the minimum number of conference rooms required.

 

Example 1:

Input: intervals = [[0,30],[5,10],[15,20]]
Output: 2

Example 2:

Input: intervals = [[7,10],[2,4]]
Output: 1

 

Constraints:

• 1 <= intervals.length <= 104
• 0 <= starti < endi <= 106

Solutions

Python3

class Solution:
    def minMeetingRooms(self, intervals: List[List[int]]) -> int:
        delta = [0] * 1000010
        for start, end in intervals:
            delta[start] += 1
            delta[end] -= 1
        for i in range(len(delta) - 1):
            delta[i + 1] += delta[i]
        return max(delta)

Java

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        int n = 1000010;
        int[] delta = new int[n];
        for (int[] e : intervals) {
            ++delta[e[0]];
            --delta[e[1]];
        }
        int res = delta[0];
        for (int i = 1; i < n; ++i) {
            delta[i] += delta[i - 1];
            res = Math.max(res, delta[i]);
        }
        return res;
    }
}

C++

class Solution {
public:
    int minMeetingRooms(vector<vector<int>>& intervals) {
        int n = 1000010;
        vector<int> delta(n);
        for (auto e : intervals) {
            ++delta[e[0]];
            --delta[e[1]];
        }
        for (int i = 0; i < n - 1; ++i) {
            delta[i + 1] += delta[i];
        }
        return *max_element(delta.begin(), delta.end());
    }
};

Go

func minMeetingRooms(intervals [][]int) int {
	n := 1000010
	delta := make([]int, n)
	for _, e := range intervals {
		delta[e[0]]++
		delta[e[1]]--
	}
	res := delta[0]
	for i := 1; i < n; i++ {
		delta[i] += delta[i-1]
		res = max(res, delta[i])
	}
	return res
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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