Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

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

这题和Insert Interval类似,但是难度要低.主要在于我们可以先对全部区间做排序.区间开头已经排完序后, 则后加入的区间b相比于开始的区间a, b.start >= a.start. 判断重合则只需要b.start <= a.end.

代码如下:

# 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]
        """
        #sweep line
        if not intervals:
            return []
        intervals.sort(key = lambda x: (x.start,x.end))
        res = [intervals[0]]
        for inter in intervals[1:]:
            if inter.start > res[-1].end: #不重合
                res.append(inter)
            else:                         #有重合
                if inter.end > res[-1].end:
                    res[-1].end = inter.end
        return res      

 

转载于:https://www.cnblogs.com/sherylwang/p/5818664.html