题目要求：数组中有一个数字出现的次数超过数组长度的一半，请找出这个数字。例如输入一个长度为9的数组{1,2,3,2,2,2,5,4,2}。由于数字2在数组中出现了5次，超过数组长度的一半，因此输出2。如果不存在则输出0。
解题思路：最简单的冒泡排序，当然不是最有答案，复杂度太大Ｏ（nk）
# -*- coding:utf-8 -*-
class Solution:
def GetLeastNumbers_Solution(self, tinput, k):
res = []
if tinput and k<=len(tinput):
for i in range(k):
jjend = len(tinput)-1
for jj in range(jjend):
if tinput[jj] < tinput[jj+1]:
tinput[jj],tinput[jj+1] = tinput[jj+1],tinput[jj]
res.append(tinput.pop())
return res
else:
return []
看到牛友把好多种排序方法都实现了，膜拜一下，先复制下来，以后看一下：
链接：https://www.nowcoder.com/questionTerminal/6a296eb82cf844ca8539b57c23e6e9bf
来源：牛客网
方法一：蒂姆排序
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# -*- coding:utf-8 -*-
class
Solution:
def
GetLeastNumbers_Solution(
self
, tinput, k):
# write code here
if
tinput
=
=
[]
or
k >
len
(tinput):
return
[]
tinput.sort()
return
tinput[: k]
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方法二：快速排序
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# -*- coding:utf-8 -*-
class
Solution:
def
GetLeastNumbers_Solution(
self
, tinput, k):
# write code here
def
quick_sort(lst):
if
not
lst:
return
[]
pivot
=
lst[
0
]
left
=
quick_sort([x
for
x
in
lst[
1
: ]
if
x < pivot])
right
=
quick_sort([x
for
x
in
lst[
1
: ]
if
x >
=
pivot])
return
left
+
[pivot]
+
right
if
tinput
=
=
[]
or
k >
len
(tinput):
return
[]
tinput
=
quick_sort(tinput)
return
tinput[: k]
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方法三：归并排序
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# -*- coding:utf-8 -*-
class
Solution:
def
GetLeastNumbers_Solution(
self
, tinput, k):
# write code here
def
merge_sort(lst):
if
len
(lst) <
=
1
:
return
lst
mid
=
len
(lst)
/
/
2
left
=
merge_sort(lst[: mid])
right
=
merge_sort(lst[mid:])
return
merge(left, right)
def
merge(left, right):
l, r, res
=
0
,
0
, []
while
l <
len
(left)
and
r <
len
(right):
if
left[l] <
=
right[r]:
res.append(left[l])
l
+
=
1
else
:
res.append(right[r])
r
+
=
1
res
+
=
left[l:]
res
+
=
right[r:]
return
res
if
tinput
=
=
[]
or
k >
len
(tinput):
return
[]
tinput
=
merge_sort(tinput)
return
tinput[: k]
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方法四：堆排序
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# -*- coding:utf-8 -*-
class
Solution:
def
GetLeastNumbers_Solution(
self
, tinput, k):
# write code here
def
siftup(lst, temp, begin, end):
if
lst
=
=
[]:
return
[]
i, j
=
begin, begin
*
2
+
1
while
j < end:
if
j
+
1
< end
and
lst[j
+
1
] > lst[j]:
j
+
=
1
elif
temp > lst[j]:
break
else
:
lst[i]
=
lst[j]
i, j
=
j,
2
*
j
+
1
lst[i]
=
temp
def
heap_sort(lst):
if
lst
=
=
[]:
return
[]
end
=
len
(lst)
for
i
in
range
((end
/
/
2
)
-
1
,
-
1
,
-
1
):
siftup(lst, lst[i], i, end)
for
i
in
range
(end
-
1
,
0
,
-
1
):
temp
=
lst[i]
lst[i]
=
lst[
0
]
siftup(lst, temp,
0
, i)
return
lst
if
tinput
=
=
[]
or
k >
len
(tinput):
return
[]
tinput
=
heap_sort(tinput)
return
tinput[: k]
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方法五：冒泡排序
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# -*- coding:utf-8 -*-
class
Solution:
def
GetLeastNumbers_Solution(
self
, tinput, k):
# write code here
def
bubble_sort(lst):
if
lst
=
=
[]:
return
[]
for
i
in
range
(
len
(lst)):
for
j
in
range
(
1
,
len
(lst)
-
i):
if
lst[j
-
1
] > lst[j]:
lst[j
-
1
], lst[j]
=
lst[j], lst[j
-
1
]
return
lst
if
tinput
=
=
[]
or
k >
len
(tinput):
return
[]
tinput
=
bubble_sort(tinput)
return
tinput[: k]
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方法六：直接选择排序
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# -*- coding:utf-8 -*-
class
Solution:
def
GetLeastNumbers_Solution(
self
, tinput, k):
# write code here
def
select_sort(lst):
if
lst
=
=
[]:
return
[]
for
i
in
range
(
len
(lst)
-
1
):
smallest
=
i
for
j
in
range
(i,
len
(lst)):
if
lst[j] < lst[smallest]:
smallest
=
j
lst[i], lst[smallest]
=
lst[smallest], lst[i]
return
lst
if
tinput
=
=
[]
or
k >
len
(tinput):
return
[]
tinput
=
select_sort(tinput)
return
tinput[: k]
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方法七：插入排序
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# -*- coding:utf-8 -*-
class
Solution:
def
GetLeastNumbers_Solution(
self
, tinput, k):
# write code here
def
Insert_sort(lst):
if
lst
=
=
[]:
return
[]
for
i
in
range
(
1
,
len
(lst)):
temp
=
lst[i]
j
=
i
while
j >
0
and
temp < lst[j
-
1
]:
lst[j]
=
lst[j
-
1
]
j
-
=
1
lst[j]
=
temp
return
lst
if
tinput
=
=
[]
or
k >
len
(tinput):
return
[]
tinput
=
Insert_sort(tinput)
return
tinput[: k]
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