题目
基本思路
这个题目算是我自己独立完成的一道题目,思路比较简单,首先是按start对intervals中的每个元素进行排序,然后依次遍历,如果发现后面的有元素的start在之前元素的[start,end]这个区间中,就把两个区间合并,同时删除第二个元素
实现代码
# 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]
"""
intervals.sort(key = lambda x:x.start)
for i,interval1 in enumerate(intervals):
for interval2 in intervals[i+1:]:
if interval1.end >= interval2.start and interval1.start <= interval2.start:
interval1.start = min(interval1.start,interval2.start)
interval1.end = max(interval1.end,interval2.end)
intervals.remove(interval2)
else:
continue
return intervals
运行结果
169 / 169 test cases passed.
Status: Accepted
Runtime: 88 ms
只击败了8.8%