问题描述:
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as
[1,2],[3,10],[12,16].This is because the new interval [4,9] overlaps with
[3,5],[6,7],[8,10].
class Solution(object):
def insert(self, intervals, newInterval):
"""
:type intervals: List[Interval]
:type newInterval: Interval
:rtype: List[Interval]
"""
start=newInterval.start
end=newInterval.end
# print('start,end',start,end)
outs=list()
index=0
flag_break=0
for index,interval in enumerate(intervals):
if interval.end>=start:
if interval.start>end:
flag_break=1
break
else:
start=min(start,interval.start)
end=max(end,interval.end)
# print('start,end:', start, end)
else:
outs.append(interval)
outs.append(Interval(start,end))
if flag_break:
outs.extend(intervals[index:])
# print(outs)
outs_num=[ (out.start,out.end) for out in outs]
return outs_num
分析:这道题的关键在于理解问题,抽取原型,理解中间可以merge部分如何界定,以及非merge部分如何进行追加list。 需要注意的是:循环到最后一个元素和在最后一个元素break的区别。
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