【Leetcode】【python】Insert Interval 插入区间

题目大意

给出多个不重合的数据区段,现在插入一个数据区段,有重合的区段要进行合并。

注意点:

所给的区段已经按照起始位置进行排序

解题思路

来自:https://shenjie1993.gitbooks.io/leetcode-python/057%20Insert%20Interval.html
最简单的方式就是复用 Merge Intervals 的方法,只需先将新的数据区段加入集合即可,但这样效率不高。既然原来的数据段是有序且不重合的,那么我们只需要找到哪些数据段与新的数据段重合,把这些数据段合并,并加上它左右的数据段即可。

代码

复用Merge Intervals

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class Solution(object):
def insert(self, intervals, newInterval):
"""
:type intervals: List[Interval]
:type newInterval: Interval
:rtype: List[Interval]
"""
result = []
if not intervals:
return [newInterval]
intervals.append(newInterval)
intervals.sort(key = lambda x: x.start)
result.append(intervals[0])
for interval in intervals[1:]:
prev = result[-1]
if prev.end >= interval.start:
prev.end = max(prev.end, interval.end)
else:
result.append(interval)
return result

独立解法(效率较高)

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

class Solution(object):
def insert(self, intervals, newInterval):
"""
:type intervals: List[Interval]
:type newInterval: Interval
:rtype: List[Interval]
"""
start, end = newInterval.start, newInterval.end
left = list(filter(lambda x: x.end < start, intervals))
right = list(filter(lambda x: x.start > end, intervals))
print left, right
if len(left) + len(right) != len(intervals): # 如果左边和右边的数量相加不等于原数量,说明新区间与某个老区间有交叉,则需要合并
start = min(start, intervals[len(left)].start) # len(left)是分开后左边区间的后面一个
end = max(end, intervals[-len(right) - 1].end) # -len(right) - 1是分开后右边区间的前面一个

return left + [Interval(start, end)] + right # Interval是区间类,新建一个区间类

总结

关于filter,list(filter(lambda x: x.end < start, intervals)),请看:
http://blog.csdn.net/shark0001/article/details/1363564