56. Merge Intervals && 57. Insert Interval
Merge Intervals
这个简单，常见的排序合并思路，然后如果前面的end大于当前的start即合并
class Solution(object):
def merge(self, intervals):
"""
:type intervals: List[Interval]
:rtype: List[Interval]
"""
if len(intervals)==0:return []
intervals.sort(key=lambda x:x.start)
ret=[]
st=intervals[0].start
ed=intervals[0].end
for idx in range(1,len(intervals)):
if ed >= intervals[idx].start:
ed=max(ed,intervals[idx].end)
else:
ret.append([st,ed])
st=intervals[idx].start
ed=intervals[idx].end
ret.append([st,ed])
return ret
Insert Interval
Insert Interval这个做了好久，分了各种情况讨论，其实转换思路成合并区间即可，如果前面的end大于当前的start即合并
class Solution:
def insert(self, intervals, newInterval):
intervals.append(newInterval)
intervals.sort(key = lambda x:x.start)
length=len(intervals)
res=[]
for i in range(length):
if res==[]:
res.append(intervals[i])
else:
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也可以思路清晰的分情况讨论完成。这个分情况那就很多了，横跨，非常小，最右最左，交叉等等。
对分情况要求很高，思路不清晰能WRONG到死。同时自己写了个简单的测试程序
class Solution(object):
def insert(self, aa, a):
"""
:type intervals: List[Interval]
:type newInterval: Interval
:rtype: List[Interval]
"""
if len(aa) == 0:
return [a]
aa.sort(key=lambda x:x.start)
st = -1
for idx,v in enumerate(aa):
# 完全大于当前无关
if v.end < a.start:
continue
# 完全包含于其他元素无关
if a.start >= v.start and a.end <= v.end:
return aa
# 完全小于当前，则会插入或者合并元素
if v.start > a.end:
if st == -1:
aa.insert(idx,a)
else:
aa=aa[0:st] + [Interval(min(a.start,aa[st].start),max(a.end,aa[st].end))] + aa[idx:]
return aa
if a.start <= v.start and a.end <= v.end:
if st == -1:
aa=aa[0:idx] + [Interval(a.start,v.end)] + aa[idx+1:]
else:
aa=aa[0:st] + [Interval(min(a.start,aa[st].start),max(a.end,v.end))] + aa[idx+1:]
return aa
#起点交叉 或者 整个横跨
if st == -1 and ((a.start >= v.start and a.end >= v.end) or (a.start <= v.start and a.end >= v.end) ):
st = idx
# 交叉情况 第一种是起点交叉（这个要记录），第二种是终点交叉
#if v.start <= a.start and v.end >= a.start:
#st=idx
# if st==-1 and v.end >= a.start:
# st=idx
# print v,a,st
# #终点交叉就能够处理了所以有等号
# if v.end >= a.end and v.start <= a.end:
# # 终点交叉时，起点没交叉
# if st == -1:
# aa=[Interval(a.start,v.end)] + aa[idx+1:]
# else:
# aa=aa[0:st] + [Interval(min(a,start,aa[st].start),max(a.end,v.end))] + aa[idx+1:]
# return aa
#起点有交叉
if st != -1:
aa=aa[0:st] + [Interval(min(a.start,aa[st].start),max(aa[st].end,a.end))]
else:
# 可能最长
if aa[0].start > a.start:
aa=[a]
else:
aa.append(a)
return aa def testOne():
p1=[[0,5],[9,12]]
p2=[7,16]
p1,p2=get_input(p1,p2)
a=Solution().insert(p1,p2)
test_data=[
([[1,5]],[0,3]),
([[1,2],[3,5],[6,7],[8,10],[12,16]],[4,9]),
([[0,5],[9,12]],[7,16]),
([[1,3],[6,9]],[2,5]),
([[1,5]],[0,6]),
([[1,5]],[0,5]),
([[1,4],[9,12],[19,22]],[7,13]),
([[3,6],[9,9],[11,13],[14,14],[16,19],[20,22],[23,25],[30,34],[41,43],[45,49]],[29,32])
]
stand_data=[
[[0,5]],
[[1,2],[3,10],[12,16]],
[[0,5],[7,16]],
[[1,5],[6,9]],
[[0,6]],
[[0,5]],
[[1,4],[7,13],[19,22]],
[[3,6],[9,9],[11,13],[14,14],[16,19],[20,22],[23,25],[29,34],[41,43],[45,49]],
]
a=map(lambda x:Solution().insert(x[0],x[1]),map(lambda x:get_input(x[0],x[1]),test_data))
#print a
#print cmp(a,stand_data)
for l1,l2 in zip(stand_data,a):
#print l1,'==>',l2
l1 = ('%s'%l1).replace(' ','')
l2 = ('%s'%l2).replace(' ','')
#print l1,l2
if l1!=l2:
print l1,'==>',l2