Merge Interval

题目描述

Given [1,3],[2,6],[8,10],[15,18], return [1,6],[8,10],[15,18].

解题方法

先按照start time sort, 然后类似insert的思想，始终维护一个interval去compare.

Solution

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

class Solution(object):
    def merge(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: List[Interval]
        """
        length = len(intervals)
        if length <= 1:
            return intervals

        intervals.sort(key=lambda x:x.start) # first sort according to start time

        result = []
        comb_interval = intervals[0]

        for i in range(1, length):
            cur_interval = intervals[i]
            if comb_interval.end >= cur_interval.start:
                comb_interval.end = max(comb_interval.end, cur_interval.end)
            else:
                result.append(comb_interval)
                comb_interval = cur_interval
        result.append(comb_interval)

        return result

Reference