36.leetCode 57:Insert Interval（插入区间）

【题目】：

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]

.

【思路】：

1. 和上一题的思路一样，先把新区间插入到列表中，使得列表仍为有序的。再遍历列表，逐一进行区间的合并

2.遍历一次列表，若列表中的当前元素与新区间没有重叠则直接插入到结果集中，否则要先与新区间进行合并，直到没有重叠为止

【Python代码】：

1.排序、合并

class Solution:
def insert(self, intervals, newInterval):
"""
:type intervals: List[Interval]
:type newInterval: Interval
:rtype: List[Interval]
"""
if not intervals:
return [newInterval]

merged = []
if len(intervals) == 1:
if newInterval.start >= intervals[0].start:
intervals.append(newInterval)
else:
intervals.insert(0, newInterval)
elif newInterval.start >= intervals[-1].start:
intervals.append(newInterval)
elif newInterval.start <= intervals[0].start:
intervals.insert(0, newInterval)
else:
for i in range(len(intervals) - 1):
if newInterval.start >= intervals[i].start and newInterval.start <= intervals[i + 1].start:
intervals.insert(i + 1, newInterval)
break

for interval in intervals:
if not merged or merged[-1].end < interval.start:
merged.append(interval)
else:
merged[-1].end = max(merged[-1].end, interval.end)

return merged

2.遍历、合并

class Solution:
def insert(self, intervals, newInterval):
"""
:type intervals: List[Interval]
:type newInterval: Interval
:rtype: List[Interval]
"""
if not intervals:
return [newInterval]

merged = []
for interval in intervals:
if newInterval is None or interval.end < newInterval.start:
merged.append(interval)
if len(intervals) == 1:
merged.append(newInterval)

elif interval.start > newInterval.end:
merged.append(newInterval)
merged.append(interval)
newInterval = None

else:
newInterval.start = min(interval.start,newInterval.start)
newInterval.end = max(interval.end,newInterval.end)
if newInterval and newInterval not in merged:
merged.append(newInterval)
return merged