# 插入区间
给你一个 无重叠的 ，按照区间起始端点排序的区间列表。
在列表中插入一个新的区间，你需要确保列表中的区间仍然有序且不重叠（如果有必要的话，可以合并区间）。
示例 1：
输入：intervals = [[1,3],[6,9]], newInterval = [2,5]
输出：[[1,5],[6,9]]
示例 2：
输入：intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
输出：[[1,2],[3,10],[12,16]]
解释：这是因为新的区间 [4,8] 与 [3,5],[6,7],[8,10] 重叠。
示例 3：
输入：intervals = [], newInterval = [5,7]
输出：[[5,7]]
示例 4：
输入：intervals = [[1,5]], newInterval = [2,3]
输出：[[1,5]]
示例 5：
输入：intervals = [[1,5]], newInterval = [2,7]
输出：[[1,7]]
提示：
0 <= intervals.length <= 104intervals[i].length == 20 <= intervals[i][0] <= intervals[i][1] <= 105intervals 根据 intervals[i][0] 按 升序 排列newInterval.length == 20 <= newInterval[0] <= newInterval[1] <= 105
## template
```python
class Interval(object):
def __init__(self, s=0, e=0):
self.start = s
self.end = e
class Solution(object):
def list2interval(self, list_interval):
ret = []
for i in list_interval:
interval = Interval(i[0], i[1])
ret.append(interval)
return ret
def interval2list(self, interval):
ret = []
x = [0,0]
for i in interval:
x[0] = i.start
x[1] = i.end
ret.append(x)
x = [0,0]
return ret
def insert(self, intervals, newInterval):
"""
:type intervals: List[Interval]
:type newInterval: Interval
:rtype: List[Interval]
"""
if intervals is None or len(intervals) == 0:
return [newInterval]
intervals = self.list2interval(intervals)
newInterval = Interval(newInterval[0], newInterval[1])
intervals.sort(key=lambda x:x.start)
pos = 0
while pos < len(intervals):
if newInterval.end < intervals[pos].start:
intervals.insert(pos, newInterval)
intervals = self.interval2list(intervals)
return intervals
if self.check_overlap(intervals[pos], newInterval):
temp = intervals.pop(pos)
newInterval = self.merge_intervals(temp, newInterval)
else:
pos += 1
if len(intervals) == 0 or pos == len(intervals):
intervals.append(newInterval)
intervals = self.interval2list(intervals)
return intervals
def check_overlap(self, curr_int, new_int):
if curr_int.start <= new_int.start:
if curr_int.end > new_int.start:
return True
else:
if curr_int.start <= new_int.end:
return True
return False
def merge_intervals(self, int1, int2):
temp_int = Interval()
temp_int.start = min([int1.start, int2.start])
temp_int.end = max([int1.end, int2.end])
return temp_int
# %%
s = Solution()
print(s.insert(intervals = [[1,3],[6,9]], newInterval = [2,5]))
```
## 答案
```python
```
## 选项
### A
```python
```
### B
```python
```
### C
```python
```