【leetcode】数组类——区间问题

数组类——区间问题

序号	题目	难度	代码
56	Merge Intervals	Medium	python、java、c++
57	Insert Interval	Hard	python、java、c++
252	Meeting Rooms	easy	python、java、c++
253	Meeting Rooms II	medium	python、java、c++
352	Data Stream as Disjoint Intervals	hard	python、java、c++

T56-Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

# Definition for an interval.
# class Interval:
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution:
    def merge(self, intervals: List[Interval]) -> List[Interval]:
        ans = []
        for intv in sorted(intervals, key=lambda x :x.start):
            if ans and ans[-1].end >= intv.start:
                ans[-1].end = max(ans[-1].end,intv.end)
            else:
                ans.append(intv)
        return ans

T57-Insert Interval

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]

Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

# Definition for an interval.
# class Interval:
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution:
    def insert(self, intervals: List[Interval], newInterval: Interval) -> List[Interval]:
        s, e = newInterval.start, newInterval.end
        left = list(filter(lambda x: x.end < newInterval.start, intervals))
        right = list(filter(lambda x: x.start > newInterval.end, intervals))
        if left + right != intervals:
            s = min(intervals[len(left)].start, s)
            e = max(intervals[~len(right)].end, e)
        return left + [Interval(s, e)] + right

T252-Meeting Rooms

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings.

Example :

For example, Given [[0, 30],[5, 10],[15, 20]], return false.

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

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

T253-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.

Example :

For example, Given [[0, 30],[5, 10],[15, 20]], return 2.

class Solution(object):
    def minMeetingRooms(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: int
        """
        meetings = []
        for i in intervals:
            meetings.append((i.start, 1))
            meetings.append((i.end, 0))
        meetings.sort()
        ans = 0
        count = 0
        for meeting in meetings:
            if meeting[1] == 1:
                count += 1
            else:
                count -= 1
            ans = max(ans, count)
        return ans

T352-Data Stream as Disjoint Intervals

Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen so far as a list of disjoint intervals.

For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be:

[1, 1]
[1, 1], [3, 3]
[1, 1], [3, 3], [7, 7]
[1, 3], [7, 7]
[1, 3], [6, 7]

Follow up:
What if there are lots of merges and the number of disjoint intervals are small compared to the data stream's size?