[leetcode] 56. Merge Intervals @ python

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.

解法

先将区间按照start升序排列, 然后遍历列表, 将每次读取的区间与结果列表里最后一个区间进行比较, 如果有重合, end取较大值, 如果没有重合, 则将区间加到结果里.

Time: O(n)
Space: O(1)

代码

# 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):
        """
        :type intervals: List[Interval]
        :rtype: List[Interval]
        """
        # edge case
        if not intervals: return []
        
        intervals.sort(key = lambda x: x.start)
        res = [intervals[0]]
        for i in range(1, len(intervals)):
            if intervals[i].start <= res[-1].end:
                res[-1].end = max(res[-1].end, intervals[i].end)
            else:
                res.append(intervals[i])
        return res