版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/u013596119/article/details/82461775
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 considerred overlapping.
Answer:
先进行排序
# 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]
"""
l=len(intervals)
if l<=1:
return intervals
intervals=sorted(intervals,key=lambda interval:interval.start)
left=intervals[0].start
right=intervals[0].end
result=[]
for i in range(1,l):
templeft=intervals[i].start
tempright=intervals[i].end
if templeft>right:
result.append(Interval(left,right))
left=templeft
right=tempright
continue
else:
if right<tempright:
right=tempright
result.append(Interval(left,right))
return result