题目来源
https://leetcode.com/problems/insert-interval/
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].
题意分析
Input:a list of intervals and a new interval
Output:insert the new interval into the list and merge if necessary
Conditions:将新的interval插入进去，如果需要合并则合并
题目思路
感觉跟merge interval很像，所以直接append list中，然后排序按照上一题merge的方法，然后过了……
AC代码（Python）
# 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]
"""
intervals.append(newInterval)
intervals.sort(key = lambda x:x.start) length = len(intervals)
res = [] if length == 0:
res.append(newInterval)
return res res.append(intervals[0])
for i in range(1,length):
size = len(res)
if res[size - 1].start <= intervals[i].start <= res[size - 1].end:
res[size - 1].end = max(intervals[i].end, res[size - 1].end)
else:
res.append(intervals[i]) return res