Skip to content

Commit

Permalink
归并排序
Browse files Browse the repository at this point in the history
  • Loading branch information
qiwsir committed Jun 23, 2014
1 parent 026ee6f commit be9e51a
Showing 1 changed file with 81 additions and 81 deletions.
162 changes: 81 additions & 81 deletions merge_sort.md
Expand Up @@ -53,84 +53,84 @@

#解决(Python)

#! /usr/bin/env python
#coding:utf-8

#方法1:将前面讲述的过程翻译过来了,略先拙笨

def merge_sort(seq):
if len(seq) ==1:
return seq
else:
middle = len(seq)/2
left = merge_sort(seq[:middle])
right = merge_sort(seq[middle:])

i = 0 #left 计数
j = 0 #right 计数
k = 0 #总计数

while i < len(left) and j < len(right):
if left[i] < right [j]:
seq[k] = left[i]
i +=1
k +=1
else:
seq[k] = right[j]
j +=1
k +=1
remain = left if i<j else right
r = i if remain ==left else j

while r<len(remain):
seq[k] = remain[r]
r +=1
k +=1

return seq

#方法2:在按照顺序取数值方面,应用了list.pop()方法,代码更紧凑简洁
#此方法来[自维基百科:归并操作](http://zh.wikipedia.org/zh/%E5%BD%92%E5%B9%B6%E6%8E%92%E5%BA%8F)

def merge_sort(lst): #此方法来自维基百科:http://zh.wikipedia.org/zh/%E5%BD%92%E5%B9%B6%E6%8E%92%E5%BA%8F
if len(lst) <= 1:
return lst

def merge(left, right):
merged = []

while left and right:
merged.append(left.pop(0) if left[0] <= right[0] else right.pop(0))

while left:
merged.append(left.pop(0))

while right:
merged.append(right.pop(0))

return merged

middle = int(len(lst) / 2)
left = merge_sort(lst[:middle])
right = merge_sort(lst[middle:])
return merge(left, right)

#方法3:原来在python的模块heapq中就提供了归并排序的方法,只要将分解后的结果导入该方法即可
#强大。
#以下方法来自[resettacode](http://rosettacode.org/wiki/Sorting_algorithms/Merge_sort#Python),并稍作修改

from heapq import merge

def merge_sort(seq):
if len(seq) <= 1:
return m
else:
middle = len(seq)/2
left = merge_sort(seq[:middle])
right = merge_sort(seq[middle:])
return list(merge(left, right)) #heapq.merge()

if __name__=="__main__":
seq = [1,3,6,2,4]
print merge_sort(seq)
#! /usr/bin/env python
#coding:utf-8
#方法1:将前面讲述的过程翻译过来了,略先拙笨
def merge_sort(seq):
if len(seq) ==1:
return seq
else:
middle = len(seq)/2
left = merge_sort(seq[:middle])
right = merge_sort(seq[middle:])
i = 0 #left 计数
j = 0 #right 计数
k = 0 #总计数
while i < len(left) and j < len(right):
if left[i] < right [j]:
seq[k] = left[i]
i +=1
k +=1
else:
seq[k] = right[j]
j +=1
k +=1
remain = left if i<j else right
r = i if remain ==left else j
while r<len(remain):
seq[k] = remain[r]
r +=1
k +=1
return seq
#方法2:在按照顺序取数值方面,应用了list.pop()方法,代码更紧凑简洁
#此方法来[自维基百科:归并操作](http://zh.wikipedia.org/zh/%E5%BD%92%E5%B9%B6%E6%8E%92%E5%BA%8F)
def merge_sort(lst): #此方法来自维基百科:http://zh.wikipedia.org/zh/%E5%BD%92%E5%B9%B6%E6%8E%92%E5%BA%8F
if len(lst) <= 1:
return lst
def merge(left, right):
merged = []
while left and right:
merged.append(left.pop(0) if left[0] <= right[0] else right.pop(0))
while left:
merged.append(left.pop(0))
while right:
merged.append(right.pop(0))
return merged
middle = int(len(lst) / 2)
left = merge_sort(lst[:middle])
right = merge_sort(lst[middle:])
return merge(left, right)
#方法3:原来在python的模块heapq中就提供了归并排序的方法,只要将分解后的结果导入该方法即可
#强大。
#以下方法来自[resettacode](http://rosettacode.org/wiki/Sorting_algorithms/Merge_sort#Python),并稍作修改
from heapq import merge
def merge_sort(seq):
if len(seq) <= 1:
return m
else:
middle = len(seq)/2
left = merge_sort(seq[:middle])
right = merge_sort(seq[middle:])
return list(merge(left, right)) #heapq.merge()
if __name__=="__main__":
seq = [1,3,6,2,4]
print merge_sort(seq)

0 comments on commit be9e51a

Please sign in to comment.