Meeting Rooms

描述

Given an array of meeting time intervals where intervals[i] = [start_i, end_i], determine if a person could attend all meetings.

Example 1:

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

Example 2:

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

分析

先排序，然后遍历数组，判断前后两个区间是否重叠。

Python

// Meeting Rooms
// Time Complexity: O(nlogn), Space Complexity: O(1)
class Solution {
    public boolean canAttendMeetings(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        for (int i = 0; i < intervals.length - 1; i++) {
            if (intervals[i][1] > intervals[i + 1][0]) {
                return false;
            }
        }
        return true;
    }
}

Java

// TODO

C++

# Meeting Rooms
# Time Complexity: O(nlogn), Space Complexity: O(1)
def canAttendMeetings(intervals):
    intervals.sort(key=lambda x: x[0])
    for i in range(len(intervals) - 1):
        if intervals[i][1] > intervals[i + 1][0]:
            return False
    return True

相关题目

• Meeting Rooms II