lintcode会议室

给定一系列的会议时间间隔，包括起始和结束时间[[s1,e1]，[s2,e2]，…(si < ei)，确定一个人是否可以参加所有会议。

第一次作答：

"""
Definition of Interval.
class Interval(object):
    def __init__(self, start, end):
        self.start = start
        self.end = end
"""

class Solution:
    """
    @param intervals: an array of meeting time intervals
    @return: if a person could attend all meetings
    """
    def canAttendMeetings(self, intervals):
        # Write your code here
        if intervals == None or len(intervals) == 0:
            return True
        
        def intervals_start(intervals):
            return intervals.start
        intervals.sort(key = intervals_start)
        for i in range(len(intervals) - 1):
            if  intervals[i + 1].start < intervals[i].end:
                return False
        return True

先对列表排序，再把相邻的元素做对比。例如第一个，若e1 > s2,则返回false。所有相邻的数据对比完后，没有返回false，则返回true

查看九章算法给的答案：
https://www.jiuzhang.com/solution/meeting-rooms/#tag-other-lang-python

class Solution(object):
    def canAttendMeetings(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: bool
        """
        if len(intervals) == 0:
            return True
        
        intervals = sorted(intervals, key = lambda x: x.start)
        
        end = intervals[0].end
        for i in range(1, len(intervals)):
            if end > intervals[i].start:
                return False
            end = intervals[i].end
        return True

发现了对对多关键字的排序。使用lambda表达式。

更简洁的答案：

class Solution:
    
    def canAttendMeetings(self, intervals) -> bool:
    
        end_time = -1
        for interval in sorted(intervals, key = lambda interval: interval.start):
            if interval.start < end_time:
                return False
            end_time = interval.end 
        return True

将循环与排序放到一起，并且用一个变量记录结束时间，然后循环开始时间。